[matrix/leetcode](sort)Search a 2D Matrix-搜索范围的缩小

落日映苍穹つ 2021-09-28 07:48 237阅读 0赞

### Serach a 2D Matrix-搜索范围的缩小 ###

*   *  题干
     *  解析
     *  代码

## 题干 ##

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.

Please submit a cpp file with the implementation of

function bool searchMatrix(vector<vector<int>>& matrix, int target)

you can start like this:

bool Solution::searchMatrix(vector<vector<int> >& matrix, int target) {}

## 解析 ##

我们需要在每一次搜索的时候可以缩小行和列的范围,可以发现当从第0,0个元素开始的时候还需要"折返"回来搜索  
技巧: 从矩阵的左下角或者右上角搜索,可以缩小行/列的范围

## 代码 ##

#include"Solution.h"
    bool Solution::searchMatrix(vector<vector<int> >& matrix, int target) {
    	int rsize=matrix.size();
    	if(rsize==0) return false;
    	int csize=matrix[0].size();
    	if(csize==0) return false;
    	if(target<matrix[0][0]) return false;
    	if(target>matrix[rsize-1][csize-1]) return false;
    	
    	int row=rsize-1;
    	int col=0;
    	//从左下角开始搜索 
    	
    	//从矩阵的四个顶点中发现，从右上角或者左下角开始搜索，就能排除行或者列 
    	while(col<= csize-1 && row>=0){
    		if(matrix[row][col]>target){
    			row--;
    		}
    		else if(matrix[row][col]<target){
    			col++;
    		}
    		else{
    			return true;
    		} 
    	}
    	return false;
    }