剑指Offer&Bigo：旋转矩阵 #57

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1. sisterAn

   added

   Bigo

   剑指Offer

   on May 31, 2020
2. 7777sea commented on Jun 1, 2020

   

   7777sea

   on Jun 1, 2020

   Last edited by sisterAn

   More actions

   思路： 按对角线反转后再逐行倒序

   [                     
       [1,2,3],     
       [4,5,6],     =>   
       [7,8,9]
   ]
   
   [
       [1,4,7],     
       [2,5,8],     =>   
       [3,6,9]
   ]
   
   [
       [7,4,1],     
       [8,5,2],     =>   
       [9,6,3]
   ]
   

   /**
    * @param {number[][]} matrix
    * @return {void} Do not return anything, modify matrix in-place instead.
    */
   var rotate = function(matrix) {
       const n = matrix.length;
       //对角线反转 0,0  n-1,n-1
       for(let i = 0; i < n; i++) {
           for(let j = 0; j < i; j++) {
               swap(matrix, [i, j], [j, i]);
           }
       }
   
       //中线左右反转
       for(let i = 0; i < n; i++) {
           for(let j = 0; j < n / 2; j++) {
               swap(matrix, [i, j], [i, n - 1 - j]);
           }
       }
   
       function swap(matrix, [x1, y1], [x2, y2]) {
           const tmp = matrix[x1][y1];
           matrix[x1][y1] = matrix[x2][y2];
           matrix[x2][y2] = tmp;
       }
   };

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3. dinjufen commented on Jun 1, 2020

   

   dinjufen

   on Jun 1, 2020

   Last edited by sisterAn

   More actions

   function rotateMatrix(mat) {
     const len = mat.length
   
     for(let i = 0;i < len;i ++) {
       for (let j = i;j < len;j ++) {
         let tmp = mat[i][j]
         mat[i][j] = mat[j][i]
         mat[j][i] = tmp
       }
     }
   
     for(let i = 0;i< len;i ++) {
       for(let j = 0;j < Math.floor(len / 2);j ++) {
         let tmp = mat[i][j]
         mat[i][j] = mat[i][len - j - 1]
         mat[i][len - j - 1] = tmp
       }
     }
   }
4. syc666 commented on Jun 1, 2020

   

   syc666

   on Jun 1, 2020

   Last edited by syc666

   More actions

   const fun = (arr) => {
       for(let i=0,len=arr.length;i<len;i++){
           for(let j=i;j<len;j++){
               const temp = arr[i][j]
               arr[i][j] = arr[j][i]
               arr[j][i] = temp
           }
       }
       arr.forEach(val=>{
           val.reverse()
       })
   }

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5. plane-hjh commented on Jun 1, 2020

   

   plane-hjh

   on Jun 1, 2020

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   const rotate = (matrix) =>{
       for(let i = 0; i < matrix.length; i++){
           for (let j = i; j < matrix[i].length; j++){
               [matrix[i][j],matrix[j][i]] = [matrix[j][i],matrix[i][j]]
           }
       }
       matrix.forEach(row=> row.reverse())
   };
6. fanefanes commented on Jun 2, 2020

   

   fanefanes

   on Jun 2, 2020

   More actions

   思路：
   es6 结构赋值，解决 不占用额外内存空间；
   通过两个指针来实现旋转；
   /**

   • @PARAM {number[][]} matrix
   • @return {void} Do not return anything, modify matrix in-place instead.
     */

   var rotate = function(matrix) {
       const n  = matrix[0].length
       for(let i = 0 ;i<=(n/2-1);i++){
           for(let j = i; j<n-i-1; j++){
               [matrix[i][j], matrix[j][n-i-1], matrix[n-i-1][n-j-1], matrix[n-j-1][i]] =  [matrix[n-j-1][i], matrix[i][j],  matrix[j][n-i-1], matrix[n-i-1][n-j-1]] 
           }
       }
       return matrix
   };
7. i7eo commented on Sep 3, 2020

   

   i7eo

   on Sep 3, 2020

   Last edited by i7eo

   More actions

   // 思路：先转置后反转

   var rotate = function(matrix) {
       var n = matrix.length
       for(let i = 0; i< n; i++) {
           // i + 1 是为了避免不必要的交换，加快速度
           for(let j = i+1; j< n; j++){
               var temp = matrix[i][j]
               matrix[i][j] = matrix[j][i]
               matrix[j][i] = temp
           }
           // 没有这一句就是矩阵转置，加上后就是旋转矩阵。反转每一行即可达到目的
           matrix[i].reverse()
       }
       return matrix
   };