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; } };