240. Search a 2D Matrix II

问题描述:

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.

 

 

解题思路:

这道题一开始我想从matrix[0][0] 即左上角开始遍历,但是后来写移动的时候发现有的地方移动不到然后还可能会出现死循环。

看了C++ with O(m+n) complexity的解法后,我发现应该从右上角开始遍历。

因为右上角的特点:

  向左变小,向下变大

左下角也有自己的特点:向上变小,向右变大

 

代码:

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

 

posted @ 2018-06-24 04:45  妖域大都督  阅读(88)  评论(0编辑  收藏  举报