第二周

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240. Search a 2D Matrix II

Description Submission Solutions

Total Accepted: 67568
Total Submissions: 177934
Difficulty: Medium
Contributors: Admin

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.

思路：

最简单的做法就是逐个搜索，时间复杂度为O(m*n)；但是因为矩阵的行和列都已排好序，所以可以利用这个特点降低搜索的复杂度。我采用的方法是从第一行的最后一个元素开始搜索，如果比target大则往该行的前一元素查找，而如果比target小则往该列的下一行元素找，当搜索到最后一行或第一列的时候，搜索结束，返回false。这种搜索方式的时间复杂度为O(m+n)。

代码：

class Solution {
public:
    bool searchMatrix(vector
     
      
       >& matrix, int target) {
        int m = matrix.size();
        if(m == 0) return false;
        int n = matrix[0].size();
        int i = 0;
        int j = n - 1;
        for(; i < m && j >= 0; ){
            if(matrix[i][j] == target) return true;
            else if(matrix[i][j] > target) j --;
            else i ++;
        }
        return false;
    }
};