LeetCode 240. Search a 2D Matrix II

题目：

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 in ascending from left to right.
• Integers in each column are sorted in ascending from top to bottom. 

For example,

Consider the following matrix:

[
  [1,   4,  7, 11, 15],
  [2,   5,  8, 12, 19],
  [3,   6,  9, 16, 22],
  [10, 13, 14, 17, 24],
  [18, 21, 23, 26, 30]
]

Given target = 5, return true.

Given target = 20, return false.

思路：

1.二分查找第一列找到最高行，二分查找最后一列找到最低行，然后对它们之间每一行进行二分查找  O(m * lgn)

2.直接对每行进行二分查找 O(m * lgn)

3.对于每一个数，上方比它小，右方比它大 O(m * n)

代码 C++ 思路2:

bool bin_search(vector<int>& num , int target){
    int start = 0;
    int end = num.size() - 1;
    
    while (start <= end) {
        int mid = start + (end - start) / 2;
        if (num[mid] == target)
            return true;
        else if (num[mid] > target)
            end = mid - 1;
        else if (num[mid] < target)
            start = mid + 1;
    }
    return false;
}

class Solution {
public:
    bool searchMatrix(vector<vector<int>>& matrix, int target) {
        for (int i = 0; i <= matrix.size() - 1; i++) {
            if (bin_search(matrix[i],target))
                return true;
        }
        return false;
    }
};

LeetCode 240. Search a 2D Matrix II

原文：http://www.cnblogs.com/gavinxing/p/5175157.html