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Commit 2a41ef1

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会员题
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import java.util.Arrays;
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public class Solution2113 {
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public int[] elementInNums(int[] nums, int[][] queries) {
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int n = nums.length;
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int q = queries.length;
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int[] ans = new int[q];
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Arrays.fill(ans, -1);
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for (int i = 0; i < q; i++) {
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int t = queries[i][0], idx = queries[i][1];
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t %= 2 * n;
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if (t < n && idx < n - t) {
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ans[i] = nums[t + idx];
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} else if (t > n && idx < t - n) {
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ans[i] = nums[idx];
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}
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}
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return ans;
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}
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}
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/*
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2113ドル. 查询删除和添加元素后的数组
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https://leetcode.cn/problems/elements-in-array-after-removing-and-replacing-elements/description/
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给你一个 下标从 0 开始 的数组 nums。一开始,在第 0 分钟,数组没有变化。此后每过一分钟,数组的 最左边 的元素将被移除,直到数组为空。然后,每过一分钟,数组的 尾部 将添加一个元素,添加的顺序和删除的顺序相同,直到数组被复原。此后上述操作无限循环进行。
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- 举个例子,如果 nums = [0, 1, 2],那么数组将按如下流程变化:[0,1,2] → [1,2] → [2] → [] → [0] → [0,1] → [0,1,2] → [1,2] → [2] → [] → [0] → [0,1] → [0,1,2] → ...
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然后给你一个长度为 n 的二维数组 queries,其中 queries[j] = [timej, indexj],表示第 j 个查询。第 j 个查询的答案定义如下:
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- 如果在时刻 timej,indexj < nums.length,那么答案是此时的 nums[indexj];
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- 如果在时刻 timej,indexj >= nums.length,那么答案是 -1。
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请返回一个长度为 n 的整数数组 ans,其中 ans[j] 为第 j 个查询的答案。
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提示:
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1 <= nums.length <= 100
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0 <= nums[i] <= 100
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n == queries.length
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1 <= n <= 10^5
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queries[j].length == 2
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0 <= timej <= 10^5
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0 <= indexj < nums.length
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时间复杂度 O(q)。
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*/
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import java.util.ArrayDeque;
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import java.util.Arrays;
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import java.util.Queue;
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public class Solution2123 {
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public int minimumOperations(int[][] grid) {
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int m = grid.length;
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int n = grid[0].length;
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int id = 1;
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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) grid[i][j] = id++;
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}
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}
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int S = id++;
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int T = id++;
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MaxFlow maxFlow = new MaxFlow(S, T);
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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] == 0) continue;
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int x = grid[i][j];
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if (i - 1 >= 0 && grid[i - 1][j] > 0) {
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int y = grid[i - 1][j];
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// 偶数到奇数连 1
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if ((i + j) % 2 == 0) maxFlow.add(x, y, 1);
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else maxFlow.add(y, x, 1);
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}
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if (j - 1 >= 0 && grid[i][j - 1] > 0) {
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int y = grid[i][j - 1];
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// 偶数到奇数连 1
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if ((i + j) % 2 == 0) maxFlow.add(x, y, 1);
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else maxFlow.add(y, x, 1);
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}
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if ((i + j) % 2 == 0) maxFlow.add(S, x, 1);
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else maxFlow.add(x, T, 1);
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}
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}
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return (int) maxFlow.maxflow();
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}
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static class MaxFlow {
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int inf = 1 << 29, N = 100010, M = 600010;
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int[] head = new int[N], ver = new int[M], edge = new int[M], Next = new int[M], d = new int[N];
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int n, m, s, t, tot;
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long maxflow;
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Queue<Integer> q;
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public MaxFlow(int s, int t) {
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this.s = s;
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this.t = t;
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this.tot = 1;
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}
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void add(int x, int y, int z) {
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ver[++tot] = y;
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edge[tot] = z;
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Next[tot] = head[x];
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head[x] = tot;
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ver[++tot] = x;
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edge[tot] = 0;
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Next[tot] = head[y];
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head[y] = tot;
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}
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boolean bfs() { // 在残量网络上构造分层图
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Arrays.fill(d, 0);
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while (!q.isEmpty()) q.remove();
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q.add(s);
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d[s] = 1;
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while (!q.isEmpty()) {
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int x = q.remove();
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for (int i = head[x]; i != 0; i = Next[i]) {
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if (edge[i] != 0 && d[ver[i]] == 0) {
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q.add(ver[i]);
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d[ver[i]] = d[x] + 1;
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if (ver[i] == t) 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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int dinic(int x, int flow) { // 在当前分层图上增广
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if (x == t) return flow;
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int rest = flow, k;
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for (int i = head[x]; i != 0 && rest != 0; i = Next[i]) {
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if (edge[i] != 0 && d[ver[i]] == d[x] + 1) {
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k = dinic(ver[i], Math.min(rest, edge[i]));
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if (k == 0) d[ver[i]] = 0; // 剪枝,去掉增广完毕的点
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edge[i] -= k;
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edge[i ^ 1] += k;
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rest -= k;
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}
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}
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return flow - rest;
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}
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long maxflow() {
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q = new ArrayDeque<>();
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long flow;
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while (bfs()) {
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while ((flow = dinic(s, inf)) != 0) maxflow += flow;
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}
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return maxflow;
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}
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}
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}
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/*
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2123ドル. 使矩阵中的 1 互不相邻的最小操作数
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https://leetcode.cn/problems/minimum-operations-to-remove-adjacent-ones-in-matrix/description/
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给你一个 下标从 0 开始 的矩阵 grid。每次操作,你可以把 grid 中的 一个 1 变成 0 。
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如果一个矩阵中,没有 1 与其它的 1 四连通(也就是说所有 1 在上下左右四个方向上不能与其他 1 相邻),那么该矩阵就是 完全独立 的。
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请返回让 grid 成为 完全独立 的矩阵的 最小操作数。
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提示:
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m == grid.length
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n == grid[i].length
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1 <= m, n <= 300
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grid[i][j] 是 0 或者 1.
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二分图最大匹配,由于 O(nm) 会 TLE。
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该用 Dinic 最大流来求。
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时间复杂度 O(sqrt(n) * m)。
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*/
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import org.junit.jupiter.api.Test;
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public class Solution2113Tests {
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private final Solution2113 solution2113 = new Solution2113();
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@Test
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public void example1() {
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}
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}
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import java.util.Arrays;
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public class Solution2604 {
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private int[] hens, grains;
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private int n, m;
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public int minimumTime(int[] hens, int[] grains) {
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Arrays.sort(hens);
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Arrays.sort(grains);
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this.hens = hens;
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this.grains = grains;
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n = hens.length;
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m = grains.length;
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int left = 0;
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int right = hens[n - 1] + grains[m - 1];
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while (left < right) {
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int mid = left + (right - left) / 2;
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// 边界二分 F, F,..., F, [T, T,..., T]
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// ----------------------^
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if (checkMid(mid)) {
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right = mid;
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} else {
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left = mid + 1;
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}
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}
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return left;
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}
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private boolean checkMid(int time) {
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int i = 0, j = 0;
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while (i < n && j < m) {
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int left = hens[i], right = hens[i];
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while (j < m) {
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left = Math.min(left, grains[j]);
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right = Math.max(right, grains[j]);
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if (Math.min(hens[i] - left, right - hens[i]) + right - left > time) break;
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j++;
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}
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i++;
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}
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return j == m;
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}
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}
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/*
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2604ドル. 吃掉所有谷子的最短时间
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https://leetcode.cn/problems/minimum-time-to-eat-all-grains/description/
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一条线上有 n 只母鸡和 m 颗谷子。给定两个整数数组 hens 和 grains ,它们的大小分别为 n 和 m ,表示母鸡和谷子的初始位置。
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如果一只母鸡和一颗谷子在同一个位置,那么这只母鸡可以吃掉这颗谷子。吃掉一颗谷子的时间可以忽略不计。一只母鸡也可以吃掉多颗谷子。
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在 1 秒钟内,一只母鸡可以向左或向右移动 1 个单位。母鸡可以同时且独立地移动。
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如果母鸡行动得当,返回吃掉所有谷子的 最短 时间。
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提示:
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1 <= hens.length, grains.length <= 2*10^4
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0 <= hens[i], grains[j] <= 10^9
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二分答案。
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*/
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public class Solution2664 {
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}
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public class Solution2907 {
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}
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public class Solution2921 {
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}
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public class Solution3073 {
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}

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