字节&剑指 Offer 29：顺时针打印矩阵（螺旋矩阵） #119

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1. sisterAn

   added

   字节

   剑指Offer

   on Oct 21, 2020
2. dinjufen commented on Oct 22, 2020

   

   dinjufen

   on Oct 22, 2020

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   用四个变量up, down, left, right分别记录当前上下左右范围，每遍历一行或一列就修改相应的值。
   例如遍历了上面第一行，就应该把up加一，遍历了右边第一列就应把right减一

   var spiralOrder = function(matrix) {
       if (matrix === undefined || matrix.length === 0) return [];
       let res = [];
       let left = 0, right = matrix[0].length-1, up = 0, down = matrix.length-1;
       while (true) {
           for (let i = left;i <= right;i++) {
               res.push(matrix[up][i]);
           }
           if (++up > down) break;
           for (let i = up;i <= down;i++) {
               res.push(matrix[i][right]);
           }
           if (--right < left) break;
           for (let i = right;i >= left;i--) {
               res.push(matrix[down][i]);
           }
           if (--down < up) break;
           for (let i = down;i >= up;i--) {
               res.push(matrix[i][left]);
           }
           if (++left > right) break;
       }
       return res;
   };

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3. U-LAN commented on Oct 22, 2020

   

   U-LAN

   on Oct 22, 2020

   Last edited by sisterAn

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   核心是四种情况的判断：

   1. 向右移动（没到右边界一直右移动遍历），到达右边界就向下移动（如果向下也是边界，说明遍历完了）
   2. 向下移动（没到下边界一直下移动遍历），到达下边界就向左移动（如果向左也是边界，说明遍历完了）
   3. 向左移动（没到左边界一直左移动遍历），到达左边界就向上移动（如果向上也是边界，说明遍历完了）
   4. 向上移动（没到上边界一直上移动遍历），到达上边界就向右移动（如果向右也是边界，说明遍历完了）

   var PrintMatrixClockwise = function (matrix) {
     let border = {      // 存储边界
       top: 0,
       left: 0,
       bottom: matrix.length - 1,
       right: matrix[0].length - 1
     } 
   
     let direction = 'right'  // top、right、bottom、left
     let result = []
     let location = [0, 0]  // 当前坐标
   
     function forword() {  // 向前移动
       switch (direction) {
         case 'top':
           if (location[1] >= border['top']) {
             // 继续向前走
             result.push(matrix[location[1]][location[0]])
             location[1]--
           } else if (location[0] + 1 <= border['right']) {
             // 此方向到头了，向右
             location[0]++
             location[1]++
             direction = 'right'
             border['left']++
           } else return
           break;
         case 'right':
           if (location[0] <= border['right']) {
             // 继续向前走
             result.push(matrix[location[1]][location[0]])
             location[0]++
           } else if (location[1] + 1 <= border['bottom']) {
             // 此方向到头了，向下
             location[0]--
             location[1]++
             direction = 'bottom'
             border['top']++
           } else return
           break;
         case 'bottom':
           if (location[1] <= border['bottom']) {
             // 继续向前走
             result.push(matrix[location[1]][location[0]])
             location[1]++
           } else if (location[0] - 1 >= border['left']) {
             // 此方向到头了，向左
             location[0]--
             location[1]--
             direction = 'left'
             border['right']--
           } else return
           break;
         case 'left':
           if (location[0] >= border['left']) {
             // 继续向前走
             result.push(matrix[location[1]][location[0]])
             location[0]--
           } else if (location[1] - 1 >= border['top']) {
             // 此方向到头了，向上
             location[0]++
             location[1]--
             direction = 'top'
             border['bottom']--
           } else return
           break;
       }
       // 递归
       forword()
     }
   
     forword()
     return result
   }
4. syc666 commented on Oct 22, 2020

   

   syc666

   on Oct 22, 2020

   Last edited by syc666

   More actions

   一圈一圈删除

   function fn(arr) {
       const res = []
       while (arr.length) {
           res.push(...arr.shift())
           if (arr.length) {
               const last = arr.pop()
               if (arr.length) {
                   arr.forEach(val => {
                       res.push(val.pop())
                   })
               }
               res.push(...last.reverse())
               if (arr.length) {
                   const temp = []
                   arr.forEach(val => {
                       temp.push(val.shift())
                   })
                   res.push(...temp.reverse())
               }
           }
       }
       return res
   }

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5. sisterAn

   added

   LeetCode

   on Dec 1, 2020
6. sisterAn

   changed the title [-]字节&剑指 Offer 29：顺时针打印矩阵[/-] [+]字节&剑指 Offer 29：顺时针打印矩阵（螺旋矩阵）[/+] on Dec 1, 2020
7. feng9017 commented on Dec 27, 2020

   

   feng9017

   on Dec 27, 2020

   More actions

   function echoValueClockWise(matrix) {
   const temp = matrix.slice();
   const result = [];

   while (temp.length > 0) {
       result.push(...temp.shift());
       temp.forEach(arr => result.push(arr.pop()));
       result.push(...(temp.pop() || []).reverse());
       for (let i = temp.length - 1; i > -1; i--) {
           result.push(temp[i].shift());
       }
   }
   
   return result;
   

   }
8. xllpiupiu commented on Oct 5, 2021

   

   xllpiupiu

   on Oct 5, 2021

   More actions

   /**
    * @param {number[][]} matrix
    * @return {number[]}
    */
   var spiralOrder = function(matrix) {
       if(matrix.length===0) {
           return []
       }
       const res = [];
       //1. 定义左边界 右边界 上边界 下边界
       let m = matrix.length;
       let n = matrix[0].length;
       let l = 0, r = n - 1, t = 0, b = m - 1;
       let matrixSize = m * n;
       while (res.length < matrixSize) {
           for (let i = l; i <= r; i++) {
               res.push(matrix[t][i]);
           }
           t++;
           for (let i = t; i <= b; i++) {
               res.push(matrix[i][r]);
           }
           r--;
           if(res.length === matrixSize) {
               break;
           }
           for (let i = r; i >= l; i--) {
               res.push(matrix[b][i]);
           }
           b--;
           for (let i = b; i >= t; i--) {
               res.push(matrix[i][l]);
           }
           l++;
       }
       return res;
   };
9. sisterAn commented on Feb 24, 2022

   

   sisterAn

   on Feb 24, 2022

   Last edited by sisterAn

   OwnerAuthor

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   var matrix = [[1,2,3,4],[5,6,7,8],[9,10,11,12]]
   
   var queryFn = function (matrix) {
     if(matrix.length===0) return []
     let res = []
     let left = 0,
       right = matrix[0].length-1,
       top = 0,
       bottom = matrix.length-1
     let n = matrix[0].length * matrix.length
     while(true) {
       if(left>right) break
       for(let i = left; i <= right; i++) {
         res.push(matrix[top][i])
       }
       top++
       if(top>bottom) break
       for(let i=top; i <= bottom; i++) {
         res.push(matrix[i][right])
       }
       right--
       if(left>right) break
       for(let i=right; i >= left; i--) {
         res.push(matrix[bottom][i])
       }
       bottom--
       if(top>bottom) break
       for(let i=bottom; i >= top; i--) {
         res.push(matrix[i][left])
       }
       left++
     }
     return res
   }
   
   queryFn(matrix)
   // [1, 2, 3, 4, 8, 12, 11, 10, 9, 5, 6, 7]