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 from left to right.
• The first integer of each row is greater than the last integer of the previous row.

 

For example,

Consider the following matrix:

[  [1,   3,  5,  7],  [10, 11, 16, 20],  [23, 30, 34, 50]]

Given target = 3, return true.

方法一，先纵向查找，再横向查找，时间复杂度为O(logm)+O(logn)

class Solution {public:    bool searchMatrix(vector
      
       
         > &matrix, int target) {        int n = matrix.size(), m = matrix[0].size();        if(target < matrix[0][0] || target >matrix[n-1][m-1]) return false;        int left = 0 , right = n-1, targetRow = 0;        //纵向二分查找        while(left <= right){            int mid = (left+right)>>1;            if(matrix[mid][0] > target ) right = mid-1;            else if(matrix[mid][0] < target) left = mid+1;            else break;        }        if(left>right) targetRow =right;        else return true;        //横向二分查找        left = 0;right = m-1;        while(left <= right){            int mid = (left+right)>>1;            if(matrix[targetRow][mid] > target) right = mid-1;            else if(matrix[targetRow][mid] < target) left = mid+1;            else break;        }        if(left <=right ) return true;        else return false;    }};
       
      

方法二，直接二分查找

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