# 240.Search a 2D Matrix II

**240.Search a 2D Matrix II**

难度：Medium

> 编写一个高效的算法来搜索 m x n 矩阵 matrix 中的一个目标值 target。该矩阵具有以下特性： 每行的元素从左到右升序排列。 每列的元素从上到下升序排列。 示例:

```
现有矩阵 matrix 如下：

[
  [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]
]
```

给定 target = 5，返回 true。 给定 target = 20，返回 false。

解决方法：因为都是有序排列，最开始想使用二分法，但是以超时告终。\
所以还是要找规律，从数组的右上角看，如果该数字比target大，则该列所有数都比target大，该列可以忽略寻找；如果该数字比target小，则该行所有数都比target小，该行也可以放弃寻找。\
从左下角开始判断也是一样的效果。\
代码如下：

```
class Solution {
public:
    bool searchMatrix(vector<vector<int>>& matrix, int target) {
        if(matrix.empty() || matrix[0].empty()) return false;
        int row=matrix.size();
        int col=matrix[0].size();
        if(matrix[row-1][col-1]<target) return false;
        int rstart=0,cstart=col-1;
        while( rstart<row && cstart >= 0)
        {
            if(matrix[rstart][cstart] == target) return true;
            else if(matrix[rstart][cstart] < target) ++rstart;
            else  --cstart;
            
        }           
        return false;
        
        
    }
};
```

在搜索判断中，如果碰到相等的就跳出循环返回`true`，搜索到左下角后，还没找到，返回`false`。


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