Search a 2D Matrix

class Solution {
public:
    bool searchMatrix(vector<vector<int> > &matrix, int target) {
        int rowNum = insertPosition(matrix, target);
        if(rowNum == -1)
        return true;
        else
        return search(matrix, target, rowNum);
    }
    
    int insertPosition(vector<vector<int> > &matrix, int target)
    {
        int rows = matrix.size();
        int left = 0;
        int right = rows - 1;
        while(left <= right)
        {
            int middle = left + ((right - left)>>1);
            if(matrix[middle][0] == target)
            return -1;
            else if(target < matrix[middle][0])
            right = middle - 1;
            else
            left = middle + 1;
        }
        if(right < 0)
        return 0;
        return right;
    }
    
    bool search(vector<vector<int> > &matrix, int target, int rowNum)
    {
        int cols = matrix[0].size();
        int left = 0;
        int right = cols - 1;
        while(left <= right)
        {
            int middle = left + ((right - left)>>1);
            if(target < matrix[rowNum][middle])
            right = middle - 1;
            else if(target > matrix[rowNum][middle])
            left = middle + 1;
            else
            return true;
        }
        return false;
    }
};

根据题意很自然的想到在纵向和横向分别使用二分查找，其中

纵向查找行号时，在没有找到元素时，需要找到比target小的最大数，也就是

其所在的行就是target有可能存在的行。另外要注意，函数insertPosition()中返回的

right，根据二分查找的规律来看就是要求的值，但对于right越界的情况需要单独处理

Search a 2D Matrix,布布扣,bubuko.com

Search a 2D Matrix

原文：http://blog.csdn.net/liyongxin4/article/details/23279225