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.
解答： 这道题的特点就是对于左下角的数而言，往上走就是变小，往右走就是变大，所以处理x，y即可。 相比于之前那道题，之前的是每一行最后一个数是小于下一行最大的，所以不一样，那道题可以两次二分法。
class Solution { public: bool searchMatrix(vector<vector<int>>& matrix, int target) { if(matrix.empty()||matrix[0].empty()) return false; int x=matrix.size()-1; int y=0; while(x>=0&&y<matrix[0].size()) { if(target==matrix[x][y]) return true; else if(target>matrix[x][y]) y++; else if(target<matrix[x][y]) x--; } return false; } };第二次是自己做出来的：
class Solution { public: bool searchMatrix(vector<vector<int>>& matrix, int target) { if(matrix.empty()||matrix[0].empty()||target>matrix.back().back()||target<matrix.front().front()) return false; int i=matrix.size()-1; int j=0; while(i>=0&&j<matrix[0].size()){ if(matrix[i][j]==target) return true; else if(matrix[i][j]<target) j++; else i--; } return false; } };