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Commit 75ec445

Browse files
20200927
1 parent cd2f5b6 commit 75ec445

22 files changed

+819
-0
lines changed

‎Java/1036.escape-a-large-maze.java

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/*
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* @lc app=leetcode id=1036 lang=java
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*
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* [1036] Escape a Large Maze
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*/
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// @lc code=start
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class Solution {
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/*
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最多200个被block, 如果走的曼哈顿距离>200说明可以
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如果target四周都被block,也不行 所以需要双向dfs
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*/
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int[] dirs = new int[]{1, 0, -1, 0, 1};
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public boolean isEscapePossible(int[][] blocked, int[] source, int[] target) {
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Set<String> blockedSet = new HashSet<>();
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for (int[] block : blocked) {
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blockedSet.add(block[0] + "->" + block[1]);
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}
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return dfs(source, target, source, blockedSet, new HashSet<>()) && dfs(target, source, target, blockedSet, new HashSet<>());
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}
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private boolean dfs(int[] source, int[] target, int[] cur, Set<String> blocked, Set<String> visited) {
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if (cur[0] == target[0] && cur[1] == target[1]) {
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return true;
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}
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if (Math.abs(source[0] - cur[0]) + Math.abs(source[1] - cur[1]) > 200) {
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return true;
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}
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for (int i = 0; i < 4; ++i) {
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int nextR = cur[0] + dirs[i];
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int nextC = cur[1] + dirs[i + 1];
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if (nextR >= 0 && nextR < 1e6 && nextC >= 0 && nextC < 1e6 && visited.add(nextR + "->" + nextC)
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&& !blocked.contains(nextR + "->" + nextC)) {
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if (dfs(source, target, new int[]{nextR, nextC}, blocked, visited)) {
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return true;
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}
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}
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}
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return false;
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}
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}
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// @lc code=end
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/*
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* @lc app=leetcode id=1162 lang=java
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*
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* [1162] As Far from Land as Possible
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*/
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// @lc code=start
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class Solution {
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/*
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判断多久1能覆盖所有0
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time: O(mn)
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space: O(mn)
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*/
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public int maxDistance(int[][] grid) {
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if (grid == null || grid.length == 0) return 0;
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int m = grid.length, n = grid[0].length;
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Queue<int[]> q = new LinkedList<>();
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for (int i = 0; i < m; ++i) {
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for (int j = 0; j < n; ++j) {
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if (grid[i][j] == 1) {
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q.offer(new int[]{i, j});
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}
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}
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}
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int[] dirs = new int[]{1, 0, -1, 0, 1};
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int step = 0;
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while (!q.isEmpty()) {
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int size = q.size();
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while (size-- > 0) {
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int[] cur = q.poll();
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for (int i = 0; i < 4; ++i) {
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int nextR = cur[0] + dirs[i];
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int nextC = cur[1] + dirs[i + 1];
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if (nextR >= 0 && nextR < m && nextC >= 0 && nextC < n && grid[nextR][nextC] == 0) {
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grid[nextR][nextC] = 1;
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q.offer(new int[]{nextR, nextC});
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}
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}
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}
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step++;
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}
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return step - 1 == 0 ? -1 : step - 1;
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}
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}
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// @lc code=end
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/*
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* @lc app=leetcode id=1218 lang=java
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*
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* [1218] Longest Arithmetic Subsequence of Given Difference
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*/
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// @lc code=start
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class Solution {
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/*
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记录以某个位置为结尾的subsequence length
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利用cur - diff的map key来查找count并更新
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time: O(n)
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space: O(n)
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*/
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public int longestSubsequence(int[] arr, int difference) {
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Map<Integer, Integer> map = new HashMap<>();
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int res = 0;
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for (int i = 0; i < arr.length; ++i) {
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map.put(arr[i], map.getOrDefault(arr[i] - difference, 0) + 1);
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res = Math.max(res, map.get(arr[i]));
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}
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return res;
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}
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}
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// @lc code=end
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‎Java/1223.dice-roll-simulation.java

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/*
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* @lc app=leetcode id=1223 lang=java
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*
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* [1223] Dice Roll Simulation
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*/
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// @lc code=start
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cclass Solution {
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/*
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dfs with memo
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dp[i][j][k] 第i次,最后一次扔出来的是j,连续了k次
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time: O(n * 6 * 6 * 15)
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space: O(n * 6 * 15)
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*/
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int[][][] dp = new int[5001][7][16];
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int kMod = (int)1e9 + 7;
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public int dieSimulator(int n, int[] rollMax) {
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if (rollMax == null || rollMax.length == 0 || n == 0) return 0;
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return dfs(rollMax, n, -1, 0);
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}
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private int dfs(int[] rollMax, int n, int last, int count) {
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if (n == 0) return 1;
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if (last >= 0 && dp[n][last][count] > 0) return dp[n][last][count];
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int res = 0;
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for (int i = 0; i < 6; ++i) {
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if (i == last && count == rollMax[i]) continue;
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res = (res + dfs(rollMax, n - 1, i, i == last ? count + 1 : 1)) % kMod;
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}
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if (last >= 0) dp[n][last][count] = res;
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return res;
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}
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}
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// @lc code=end
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/*
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* @lc app=leetcode id=1352 lang=java
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*
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* [1352] Product of the Last K Numbers
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*/
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// @lc code=start
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class ProductOfNumbers {
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/*
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prefix array
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如果有0,则将之前的清空
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如果k > size,说明这部分有0
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*/
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List<Integer> prefixProduct;
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public ProductOfNumbers() {
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prefixProduct = new ArrayList<>();
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prefixProduct.add(1);
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}
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public void add(int num) {
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if (num > 0) {
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prefixProduct.add(prefixProduct.get(prefixProduct.size() - 1) * num);
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} else {
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prefixProduct = new ArrayList<>();
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prefixProduct.add(1);
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}
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}
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public int getProduct(int k) {
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int size = prefixProduct.size();
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return size > k ? prefixProduct.get(size - 1) / prefixProduct.get(size - k - 1) : 0;
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}
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}
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/**
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* Your ProductOfNumbers object will be instantiated and called as such:
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* ProductOfNumbers obj = new ProductOfNumbers();
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* obj.add(num);
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* int param_2 = obj.getProduct(k);
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*/
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/**
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* Your ProductOfNumbers object will be instantiated and called as such:
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* ProductOfNumbers obj = new ProductOfNumbers();
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* obj.add(num);
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* int param_2 = obj.getProduct(k);
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*/
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// @lc code=end
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/*
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* @lc app=leetcode id=1423 lang=java
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*
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* [1423] Maximum Points You Can Obtain from Cards
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*/
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// @lc code=start
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class Solution {
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/*
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find min subarray with length n - k
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time: O(n)
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space: O(1)
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*/
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public int maxScore(int[] cardPoints, int k) {
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if (cardPoints == null || cardPoints.length == 0) return 0;
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int n = cardPoints.length;
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int total = 0;
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int cur = 0;
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int min = Integer.MAX_VALUE;
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for (int l = 0, r = 0; r < n; ++r) {
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total += cardPoints[r];
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cur += cardPoints[r];
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if (r - l + 1 == (n - k)) {
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min = Math.min(min, cur);
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cur -= cardPoints[l++];
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}
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}
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return min == Integer.MAX_VALUE ? total : total - min;
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}
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}
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// @lc code=end
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/*
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* @lc app=leetcode id=1477 lang=java
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*
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* [1477] Find Two Non-overlapping Sub-arrays Each With Target Sum
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*/
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// @lc code=start
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class Solution {
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/*
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建立prefix sum的map
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同时找sum - target 和sum + target
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time: O(n)
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space: O(n)
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*/
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public int minSumOfLengths(int[] arr, int target) {
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if (arr == null || arr.length == 0) return -1;
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Map<Integer, Integer> map = new HashMap<>();
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int sum = 0;
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map.put(0, -1);
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for (int i = 0; i < arr.length; ++i) {
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sum += arr[i];
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map.put(sum, i);
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}
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sum = 0;
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int leftMin = Integer.MAX_VALUE, res = Integer.MAX_VALUE;
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for (int i = 0; i < arr.length; ++i) {
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sum += arr[i];
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if (map.containsKey(sum - target)) {
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leftMin = Math.min(leftMin, i - map.get(sum - target));
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}
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if (map.containsKey(sum + target) && leftMin < Integer.MAX_VALUE) {
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res = Math.min(res, map.get(sum + target) - i + leftMin);
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}
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}
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return res == Integer.MAX_VALUE ? -1: res;
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}
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}
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// @lc code=end
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‎Java/397.integer-replacement.java

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/*
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* @lc app=leetcode id=397 lang=java
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*
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* [397] Integer Replacement
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*/
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// @lc code=start
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class Solution {
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/*
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末位为0,除以2
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末位为11,加一, 01则减1
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time: O(bitCount)
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space: O(1)
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*/
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public int integerReplacement(int n) {
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int count = 0;
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while (n != 1) {
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if ((n & 1) == 0) {
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n >>>= 1;
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} else if ((((n >>> 1) & 1) == 0) || n == 3) {
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n--;
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} else {
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n++;
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}
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count++;
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}
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return count;
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}
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}
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// @lc code=end
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‎Java/415.add-strings.java

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/*
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* @lc app=leetcode id=415 lang=java
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*
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* [415] Add Strings
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*/
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// @lc code=start
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class Solution {
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public String addStrings(String num1, String num2) {
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StringBuilder sb = new StringBuilder();
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int sum = 0;
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int i = num1.length() - 1, j = num2.length() - 1;
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while (i >= 0 || j >= 0) {
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if (i >= 0) {
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sum += num1.charAt(i) - '0';
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}
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if (j >= 0) {
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sum += num2.charAt(j) - '0';
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}
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--i;
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--j;
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sb.append(sum % 10);
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sum /= 10;
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}
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if (sum > 0) sb.append(sum);
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return sb.reverse().toString();
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}
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}
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// @lc code=end
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‎Java/418.sentence-screen-fitting.java

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/*
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* @lc app=leetcode id=418 lang=java
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*
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* [418] Sentence Screen Fitting
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*/
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// @lc code=start
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class Solution {
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/*
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一行一行的加, 如果碰到当前位置不满足末行条件,将count退格代表一部分字符不能加在这一行得加在下一行
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a _ b c d _ e _
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0 1 0 -1 -2 1 0 1
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第一次加count = 6, 下一行开始的是e, 所以之前所有的都可以放进这一行(如果是空格就补一个空格)
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第二次count = 12, 下一行开始的是d, 所以要退两格用空格补充,下一行开头的是bcd count = 12 - 2 = 10
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第三次count = 16, 下一行开始是a, 满足
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time: O(len + cols)
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sapce: O(len)
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*/
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public int wordsTyping(String[] sentence, int rows, int cols) {
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String s = String.join(" ", sentence) + " ";
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int len = s.length(), count = 0;
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int[] map = new int[len];
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for (int i = 1; i < len; ++i) {
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map[i] = s.charAt(i) == ' ' ? 1 : map[i-1] - 1;
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}
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for (int i = 0; i < rows; ++i) {
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count += cols;
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count += map[count % len];
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}
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return count / len;
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}
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}
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// @lc code=end
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