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Commit 25cb3d6

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[Function add]
1. Add leetcode solutions with amazon tag.
1 parent a56ebc3 commit 25cb3d6

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‎leetcode/273. Integer to English Words.md

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@@ -69,3 +69,52 @@ Output: "One Billion Two Hundred Thirty Four Million Five Hundred Sixty Seven Th
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}
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}
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```
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### Amazon Session
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* Method 1: Math
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```Java
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class Solution {
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private static final String[] units = new String[]{"", "Thousand", "Million", "Billion"};
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private static final String[] ones = new String[]{"", "One", "Two", "Three", "Four", "Five", "Six", "Seven", "Eight", "Nine"};
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private static String[] tens = new String[]{"", "Ten", "Twenty", "Thirty", "Forty", "Fifty", "Sixty", "Seventy", "Eighty", "Ninety"};
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private static String[] overTens = new String[]{"", "Eleven", "Twelve", "Thirteen", "Fourteen", "Fifteen", "Sixteen", "Seventeen", "Eighteen", "Nineteen"};
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public String numberToWords(int num) {
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if(num == 0) return "Zero";
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int bigUnit = 0;
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String result = "";
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while(num != 0){
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String cur = parseThree(num % 1000);
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if(cur.length() != 0){
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result = cur
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+ (bigUnit == 0 ? "": " ")
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+ units[bigUnit]
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+ (result.length() == 0 ? "": " ")
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+ result;
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}
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bigUnit++;
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num /= 1000;
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}
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return result;
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}
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private String parseThree(int num){
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String result = "";
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if(num % 100 > 10 && num % 100 < 20){
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result = overTens[num % 100 - 10];
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num /= 100;
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}else{
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if(num % 10 != 0){
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result = ones[num % 10];
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}
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num /= 10;
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if(num % 10 != 0){
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result = tens[num % 10] + (result.length() == 0 ? "" : " ") + result;
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}
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num /= 10;
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}
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if(num % 10 != 0){
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result = ones[num % 10] + " Hundred" + (result.length() == 0 ? "": " ") + result;
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}
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return result;
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}
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}
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```

‎leetcode/348. Design Tic-Tac-Toe.md

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@@ -130,3 +130,116 @@ Could you do better than O(n2) per move() operation?
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* int param_1 = obj.move(row,col,player);
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*/
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```
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### Amazon Session
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* Method 1: Design
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```Java
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class TicTacToe {
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private int[][] g;
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private int n;
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/** Initialize your data structure here. */
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public TicTacToe(int n) {
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this.g = new int[n][n];
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this.n = n;
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}
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/** Player {player} makes a move at ({row}, {col}).
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@param row The row of the board.
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@param col The column of the board.
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@param player The player, can be either 1 or 2.
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@return The current winning condition, can be either:
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0: No one wins.
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1: Player 1 wins.
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2: Player 2 wins. */
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public int move(int row, int col, int player) {
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g[row][col] = player;
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if(checkHorizontal(row, col, player) || checkVerticle(row, col, player) || checkDiagnonal(row, col, player)) return player;
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return 0;
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}
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private boolean checkHorizontal(int row, int col, int player){
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int temp = col - 1;
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while(temp >= 0){
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if(g[row][temp--] != player) return false;
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}
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temp = col + 1;
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while(temp < n){
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if(g[row][temp++] != player) return false;
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}
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return true;
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}
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private boolean checkVerticle(int row, int col, int player){
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int temp = row - 1;
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while(temp >= 0){
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if(g[temp--][col] != player) return false;
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}
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temp = row + 1;
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while(temp < n){
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if(g[temp++][col] != player) return false;
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}
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return true;
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}
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private boolean checkDiagnonal(int row, int col, int player){
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int tx = 0, ty = 0;
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boolean flag1 = true;
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while(tx < n){
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flag1 &= g[tx][tx++] == player;
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}
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tx = 0; ty = n - 1;
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boolean flag2 = true;
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while(tx < n){
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flag2 &= g[tx++][ty--] == player;
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}
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return flag1 || flag2;
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}
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}
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/**
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* Your TicTacToe object will be instantiated and called as such:
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* TicTacToe obj = new TicTacToe(n);
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* int param_1 = obj.move(row,col,player);
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*/
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```
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* Method 2:
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1. We don't have to record all of the actions.
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2. We only need to record all the numebrs in one row or col.
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```Java
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class TicTacToe {
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private int[][] rows;
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private int[][] cols;
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private int[][] digs;
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private int n;
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/** Initialize your data structure here. */
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public TicTacToe(int n) {
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this.rows = new int[n][2];
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this.cols = new int[n][2];
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this.digs = new int[2][2];
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this.n = n;
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}
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/** Player {player} makes a move at ({row}, {col}).
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@param row The row of the board.
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@param col The column of the board.
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@param player The player, can be either 1 or 2.
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@return The current winning condition, can be either:
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0: No one wins.
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1: Player 1 wins.
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2: Player 2 wins. */
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public int move(int row, int col, int player) {
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if(++rows[row][player - 1] == n) return player;
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if(++cols[col][player - 1] == n) return player;
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if(row == col && ++digs[0][player - 1] == n) return player;
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if(row + col == n - 1 && ++digs[1][player - 1] == n) return player;
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return 0;
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}
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}
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/**
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* Your TicTacToe object will be instantiated and called as such:
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* TicTacToe obj = new TicTacToe(n);
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* int param_1 = obj.move(row,col,player);
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*/
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```
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### Reference
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1. [Java concise O(1) easy understand](https://leetcode.com/problems/design-tic-tac-toe/discuss/296370/Java-concise-O(1)-easy-understand)

‎leetcode/675. Cut Off Trees for Golf Event.md

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@@ -111,5 +111,64 @@ Hint: size of the given matrix will not exceed 50x50.
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}
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```
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### Amazon Session
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* Method 1: PriorityQueue + bfs
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```Java
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class Solution {
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private int[][] g;
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private static final int[][] dir = new int[][]{{0, 1}, {0, -1}, {1, 0}, {-1, 0}};
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public int cutOffTree(List<List<Integer>> forest) {
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if(forest.get(0).get(0) == 0) return -1;
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int height = forest.size(), width = forest.get(0).size();
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//a[0]: value, a[1]: row, a[2]: col
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PriorityQueue<int[]> pq = new PriorityQueue<>((a, b)->{
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return a[0] - b[0];
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});
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g = new int[height][width];
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for(int i = 0; i < height; i++){
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for(int j = 0; j < width; j++){
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g[i][j] = forest.get(i).get(j);
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if(g[i][j] != 0){
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pq.offer(new int[]{g[i][j], i, j});
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}
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}
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}
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int[] last = new int[]{0, 0};
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int result = 0;
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LABEL:
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while(!pq.isEmpty()){
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int[] dest = pq.poll();
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Queue<int[]> q = new LinkedList<>();
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Set<Integer> visited = new HashSet<>();
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visited.add(last[0] * width + last[1]);
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q.offer(last);
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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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for(int sz = 0; sz < size; sz++){
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int[] cur = q.poll();
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if(cur[0] == dest[1] && cur[1] == dest[2]){
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result += step;
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last = cur;
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continue LABEL;
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}
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int tx = 0, ty = 0;
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for(int d = 0; d < 4; d++){
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tx = cur[0] + dir[d][0];
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ty = cur[1] + dir[d][1];
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if(tx >= 0 && tx < height && ty >= 0 && ty < width && !visited.contains(tx * width + ty) && g[tx][ty] != 0){
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visited.add(tx * width + ty);
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q.offer(new int[]{tx, ty});
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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 -1;
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}
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return result;
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}
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}
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```
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### Reference
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1. [花花酱 LeetCode 675. Cut Off Trees for Golf Event - 刷题找工作 EP55](https://www.youtube.com/watch?v=OFkLC30OxXM)

‎leetcode/909. Snakes and Ladders.md

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}
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}
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```
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### Amazon Session
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* Method 1: BFS(because it is shortest path question)
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```Java
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class Solution {
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private int N;
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private int[] numberToIndex(int num){ // return the row and col index of current number
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num--; // change to 0-based
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int diff = num / N;
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int rem = num % N;
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return new int[]{N - 1 - num / N, diff % 2 == 0 ? rem: N - 1 - rem};
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}
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public int snakesAndLadders(int[][] board) {
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this.N = board.length;
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int dest = N * N;
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Queue<Integer> q = new LinkedList<>();
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q.offer(1);
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int step = 0;
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Set<Integer> visited = new HashSet<>();
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while(!q.isEmpty()){
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int sz = q.size();
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for(int p = 0; p < sz; p++){
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int cur = q.poll();
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if(cur == dest) return step;
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for(int i = 1; i <= 6; i++){
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int next = cur + i;
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if(visited.contains(next) || next > dest) continue;
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visited.add(next);
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int[] index = numberToIndex(next);
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if(board[index[0]][index[1]] == -1) q.offer(next);
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else q.offer(board[index[0]][index[1]]);
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}
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}
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step++;
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}
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return -1;
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}
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}
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```

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