LeetCode--Search a 2D Matrix II(分治策略)

Search a 2D Matrix II

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题目

Write an efficient algorithm that searches for a value in anmxnmatrix. 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.

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, returntrue.

Given target =20, returnfalse.

分析

这里不可能用直接搜索的办法去做，根据矩阵是有序的，根据这个特点，首先想到的是二分查找，本来实现的算法是对每一行都进行二分查找，这样便将一个问题分开为若干个子问题。然后我们发现某些行可以之间去掉，因为是有序的，当受元素大于target或者末尾元素小于target时我们可以直接跳过。

源码

class Solution { public: bool searchMatrix(vector<vector<int> >& matrix, int target) { if(matrix.size() == 0||(matrix.size() != 0&&matrix[0].size() == 0)) { return false; } for(int i = 0; i < matrix.size(); i++) { if(matrix[i][matrix[0].size() - 1] < target||matrix[i][0] > target) { continue; } if(binary_search(matrix[i],target)) { return true; } } return false; } bool binary_search(vector<int> m, int target) { int begin = 0, end = m.size() - 1, middle = (begin + end)/2; while(begin <= end) { middle = (begin + end)/2; if(target == m[middle]) { return true; } else if(target < m[middle]) { end = middle -1; } else { begin = middle + 1; } } return false; } };

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