【LeetCode】【矩陣旋轉】Rotate Image

描述

You are given an n x n 2D matrix representing an image.

Rotate the image by 90 degrees (clockwise).

Note:

You have to rotate the image in-place, which means you have to modify the input 2D matrix directly. DO NOT allocate another 2D matrix and do the rotation.

Example 1:

Given input matrix = 
[
  [1,2,3],
  [4,5,6],
  [7,8,9]
],

rotate the input matrix in-place such that it becomes:
[
  [7,4,1],
  [8,5,2],
  [9,6,3]
]

Example 2:

Given input matrix =
[
  [ 5, 1, 9,11],
  [ 2, 4, 8,10],
  [13, 3, 6, 7],
  [15,14,12,16]
], 

rotate the input matrix in-place such that it becomes:
[
  [15,13, 2, 5],
  [14, 3, 4, 1],
  [12, 6, 8, 9],
  [16, 7,10,11]
]

 

思路：移位法

將矩陣的的邊框連線

如果矩陣的寬度和高度都為n，那么一共有n/2個連線框，這些連線框位移形成旋轉矩陣，如下圖 

 

 按順時針位移，依次遍歷連線上的點就行。

這里為了不增加額外的空間消耗，不創建新的矩陣，我們用一個temp變量，存放臨時的待替換元素，如果要順時針替換，則需要兩個臨時變量還要不斷更新，所以我們采用逆時針旋轉替換，這樣只需要在開始的時候加入一個temp變量即可。

然后遍歷的時候令i為旋轉的圈數，j為直線上遍歷的序號。使得j=i，j增大到n-1-i，然后根據四個點的序號關聯完成轉移。

 

class Solution {
public:
    void rotate(vector<vector<int>>& matrix) {
        int n = matrix.size();
        int count = n/2,temp = 0;
        for(int i = 0;i < count;++i){
            for(int j = i;j<n-i-1;++j){
                temp = matrix[i][j];
                matrix[i][j] = matrix[n - j - 1][i];
                matrix[n - j - 1][i] = matrix[n - i - 1][n - j - 1];
                matrix[n - i - 1][n - j - 1] = matrix[j][n - i - 1];
                matrix[j][n - i - 1] = temp;
            }
        }
    }
};