Leetcode 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.

给一个在两个维度上都递增的矩阵，判断矩阵中是否存在一个数。

如果用二分需要遍历每一行，对每一行二分判断target是否存在，复杂度为O(n*log(m))。

如果从矩阵的右上角或者左下角开始，则相邻的两个方向可以构成一个决策集，一个方向一定比当前数小，一个方向一定比当前数大，用这种方法复杂度为O(m + n).

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