題目來(lái)源
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.
看到這道題一開(kāi)始的想法就是二分搜索，先從左往右二分，再?gòu)纳贤露?，時(shí)間復(fù)雜度是O(log(m)log(n))。
然后寫(xiě)好了之后才認(rèn)清題目…發(fā)現(xiàn)根本完全就是錯(cuò)誤的…
下次記得好好看題目！??！
沒(méi)啥好想法，每一行或者每一列都二分搜索一下…
代碼如下：

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

可以AC，不過(guò)比較慢，不對(duì)，是巨慢==。
想了想怎么優(yōu)化，可以把行首和列首大于target的行和列去掉。
但是復(fù)雜度還是一樣的。然后我又看了答案。下面是復(fù)雜度O(m+n)的方法。

class Solution {
public:
    bool searchMatrix(vector<vector<int>>& matrix, int target) {
        auto m = matrix.size();
        if (m == 0)
            return false;
        auto n = matrix[0].size();
        if (n == 0)
            return false;
        int x = 0, y = n - 1;
        while (x < m && y >= 0) {
            if (matrix[x][y] == target)
                return true;
            else if (matrix[x][y] < target)
                x++;
            else if (matrix[x][y] > target)
                y--;
        }
        return false;
    }
};

這種題目要好好想想如何利用好題目的條件。