搜索二维矩阵

这道题乍一看，和做过的搜索二维矩阵Ⅱ类似，用之前的代码也能通过，但忽略掉了每行的第一个整数大于前一行的最后一个整数这个条件。可以使用两次二分法来解决这道题目

class Solution {
public:
    bool searchMatrix(vector<vector<int>>& matrix, int target) {
        int m = matrix.size() - 1, n = matrix[0].size() - 1;
        int left = 0, right = m, row = 0;
        if(target == matrix[left][n] || target == matrix[right][n]){
            return true;
        }
        while(left < right){
            int mid = left + (right - left)/2;
            if(target < matrix[mid][n]){
                right = mid;
            }else if(target > matrix[mid][n]){
                left = mid + 1;
                row = left;
            }else{
                return true;
            }
        }
        int newleft = 0, newright = n;
        while(newleft < newright){
            int newmid = newleft + (newright - newleft)/2;
            if(target < matrix[row][newmid]){
                newright = newmid;
            }else if(target > matrix[row][newmid]){
                newleft = newmid + 1;
            }else{
                return true;
            }
        }
        return false;
    }
};