coding

效能很差 /** * @param {character[][]} board * @return {boolean} */ var isValidSudoku = function(board) { var check = function(ar){ var ob = {}; for(var j=0;j<9;j++){ if(ar[j] == '.'){ continue; } if(ob[ar[j]] === undefined){ ob[ar[j]] = 1; }else{ return false; } } return true; }; var getAr = function(i){ var r = parseInt(i/3); var c = i%3; var ar = []; ar.push(board[r*3+0][c*3+0]); ar.push(board[r*3+0][c*3+1]); ar.push(board[r*3+0][c*3+2]); ar.push(board[r*3+1][c*3+0]); ar.push(board[r*3+1][c*3+1]); ar.push(board[r*3+1][c*3+2]); ar.push(board[r*3+2][c*3+0]); ar.push(board[r*3+2][c*3+1]); ar.push(board[r*3+2][c*3+2]); return ar; }; var getcAr = function(i){ var ar = []; ar.pus...

這幾天困了我很久的ㄧ題 /**  * @param {string} s  * @return {number}  */ var longestValidParentheses = function(s) {          var i = 0;     var max = 0;     var stack = [];     var stack_ind = [];          while(i&amp;lt;=s.length-1){         var pop = 0;         if(s[i] == ')'){             if(stack.length &amp;gt; 0){                 pop = stack.pop();                 stack_ind[pop] = true;                 stack_ind[i] = true;             }else{                 stack_ind[i] = false;             }             i++;       ...

/**  * @param {string} s  * @param {string[]} words  * @return {number[]}  */ var findSubstring = function(s, words) {           var counts = {};               //counts 記錄所有可用的pattern剩餘使用次數         for (var p in words){             counts[words[p]] = counts[words[p]] === undefined ? 1 : counts[words[p]] + 1;         }               var n = s.length, num = words.length, len = words[0].length;         var indexes = [];         //從頭開始比對 是否有可用pattern可以使用         //例如 s = abdegdsadf, counts= {'abd': 2, 'acc': 1}         //發現abd is match, 接下來就會檢察egd有沒有再counts裡面         //發現沒有就進入下一round, s = bdegdsadf, 開始檢察bde有沒有在count裡面         //每一round檢查完會看可用pattern是不是都用完了, 用完了代表成功match ...