LeetCode OJ | Search a 2D Matrix

LeetCode的这个问题可用“夹逼”的思想完成。
题目如下
Write an efficient algorithm that searches for a value in an m x n matrix. This matrix has the following properties:

Integers in each row are sorted in ascending from left to right.
Integers in each column are sorted in ascending from top to bottom.
For example,

Consider the following matrix:

[
[1, 4, 7, 11, 15],
[2, 5, 8, 12, 19],
[3, 6, 9, 16, 22],
[10, 13, 14, 17, 24],
[18, 21, 23, 26, 30]
]
Given target = 5, return true.

Given target = 20, return false.

问题在于怎么将搜索范围缩小。注意到每行每列数字的特点是向右、向下递增，也就是说，每个“方块”内的数，左上角的数最小，右下角的数最大，那么只要利用这点，就可以将某些“方块”排除出搜索范围。最终将搜索范围缩小到我们可以接受的范围。

附上代码：

class Solution {public:    bool flag = true;    bool searchMatrix(vector<vector<int>>& matrix, int target)     {        int left = 0,right = matrix[0].size()-1, up = 0, down = matrix.size()-1;        int left1 = -1,right1 = -1,up1 = -1,down1 = -1;        int flag = 0;        while(1)        {            left1 = left;            right1 = right;            up1 = up;            down1 = down;            flag = shrink(matrix,left,right,up,down,target);            if(left1==left&&right1==right&&up1==up&&down1==down) break;            if(flag == 1) return true;        }        for(int i=up;i<=down;++i)        for(int j=left;j<=right;++j)        if(matrix[i][j] == target) return true;        return false;    }    int shrink(vector<vector<int>>& a,int &left,int &right,int &up,int &down,int target)    {        int mid = 0;        for(int i=up;i<=down;++i)        {            if(a[i][right] > target)            {                up = i;                break;            }            if(a[i][right] == target)            return 1;            else if(i==down) up = i;        }        for(int i=down;i>=up;--i)        {            if(a[i][left] < target)            {                down = i;                break;            }            if(a[i][left] == target)            return 1;            else if(i==up) down = up;        }        for(int i=left;i<=right;++i)        {            if(a[down][i] > target)            {                left = i;                break;            }            if(a[down][i] == target) return 1;            else if(i==right) left = i;        }        for(int i=right;i>=left;--i)        {            if(a[up][i] < target)            {                right = i;                break;            }            if(a[up][i] == target) return 1;            else if(i==left) right = i;        }        return 0;    }};