LeetCode算法题目：Search a 2D Matrix II

题目：

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.

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分析：

此题与Search 2D Matrix基本相同，但现在下一列的最小值不再大于上一列的最大值，故可以采用从左下角或者右上角开始搜索的方法。

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代码：

//从左下角开始查找，也可以从右上角开始查找 class Solution { public: bool searchMatrix(vector<vector<int>>& matrix, int target) { if (matrix.empty()) return false; int n = matrix.size(); int m = matrix.front().size(); int i = n-1; int j = 0; while(i>=0 && j<m){ if (matrix[i][j] == target) return true; else if (matrix[i][j]<target) j++; else i--; } return false; } }; //从右上角开始查找 class Solution { public: bool searchMatrix(vector<vector<int>>& matrix, int target) { if(matrix.empty()) return false; int m = matrix.size(); int n = matrix.front().size(); int i=0, j=n-1; while (i<m && j>=0) { int x = matrix[i][j]; if (x == target) return true; else if (x >target) j--; else i++; } return false; } };