Search a 2D Matrix——两度二分查找

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.

这题很简单

思路：二分法确定target可能在第几行出现。再用二分法在该行确定target可能出现的位置。时间复杂度O(logn+logm)

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