搜索二维矩阵

 

 

 

详细思路

二分，left=0，right=n*m-1,mid=left+(right-left)/2，将mid映射为ij，j=mid%m; i=mid/m

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

踩过的坑

        while(left<=right){

left等于right一般要判断，否则就是等于直接离开