［LeetCode]Search a 2D Matrix II

最新推荐文章于 2022-08-05 12:18:43 发布

原创最新推荐文章于 2022-08-05 12:18:43 发布 · 469 阅读

0 ·

CC 4.0 BY-SA版权

LeetCode 专栏收录该内容

149 篇文章

订阅专栏

本文介绍了一种高效的算法，用于在一个特殊格式的二维矩阵中查找特定值。该矩阵的特点是每行和每列都按升序排列。文章提供了两种解决方案：一种是在O(m+n)的时间复杂度下，从矩阵的右上角开始逐步搜索；另一种是在O(mlogn)的时间复杂度下，对矩阵的每一行执行二分查找。

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］O（m+n）复杂度，从右上开始搜寻。由于矩阵是有序的，所以可以根据大小关系缩小范围。

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

O（mlogn)对每一行进行二分查找。

class Solution {
public:
    bool searchMatrix(vector<vector<int>>& matrix, int target) {
        int m = matrix.size();
        int n = matrix[0].size();
        for(int i=0 ; i<m ; ++i){
            if(matrix[i][0]<=target && matrix[i][n-1]>=target){
                if(searchVector(matrix[i],target))
                    return true;
            }
        }
        return false;
    }
    
   bool searchVector(vector<int>& v, int target) {
    int left = 0, right = v.size() - 1;
    while (left <= right) {
        int mid = left + (right - left) / 2;
        if (v[mid] == target)
            return true;
        if (v[mid] < target)
            left = mid + 1;
        else
         right = mid - 1;
        }

        return false;
    }
    
};