Leetcode 74. Search a 2D Matrix

74. Search a 2D Matrix

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.

题目链接：https://leetcode.com/problems/search-a-2d-matrix/description/

题目大意：在一个n * m的矩阵中查找目标数字target，找到返回true，否则返回false。矩阵满足以下条件：

1. 每行升序有序。

2. 每行第一个数大于上一行的最后一个数。

解题思路：因为矩阵的行和列有序性，可以用二分查找的方法把时间复杂度降到 log(nm) ，先二分查到所在的行，用时log(n)，再对该行二分查找，用时log(m)。

代码如下：

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