Search a 2D Matrix II(分治策略)

2019年11月23日 105次阅读

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.

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.

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;
    }
};

 

点赞