240. Search a 2D Matrix II

240. Search a 2D Matrix II

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.

分析：
首先看到题目的设定，是行和列分别降序，那么最自然的想法就是，我先在行找，如果找到一个值，它的前一个值比pivot小，后一个值比pivot大，那么再从这一列往下找，如果找到了比该值大的值还没找到，则断定没有这个值。
但是这样的做法是时间复杂度是O(n+m)，因此，对于这种有序数组中查找数的题目，我们很自然可以想到用二分法查找，用二分法代替上面的逐个查找，那么复杂的就能降到O(mlogn)。

#include <iostream>#include <vector>using namespace std;class Solution {public:    bool searchMatrix(vector<vector<int>>& matrix, int target) {        if (matrix.empty() || matrix.size() < 1 || matrix[0].size() < 1) {            return false;        }        return searchRecCol(matrix, target, 0, matrix.size() - 1);    }    bool searchRecCol(vector<vector<int>>& matrix, int target, int top, int bottom) {        if (top > bottom)            return false;        int mid = top + (bottom - top) / 2;        if (matrix[mid].front() <= target && target <= matrix[mid].back())            if (searchRow(matrix[mid], target))                return true;        if (searchRecCol(matrix, target, top, mid - 1)) return true;        if (searchRecCol(matrix, target, mid + 1, bottom)) return true;        return false;    }    bool searchRow(vector<int>& vector, int target) {        int left = 0, right = vector.size() - 1;        while (left <= right) {            int mid = left + (right - left) / 2;            if (vector[mid] == target)                return true;            if (vector[mid] < target)                left = mid + 1;            else                right = mid - 1;        }        return false;    }};