LeetCode--search-a-2d-matrix

矩阵搜索算法

最新推荐文章于 2020-09-29 20:40:03 发布

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本文介绍了一种高效的矩阵搜索算法，用于在一特殊性质的二维矩阵中查找特定值。该矩阵的每一行元素从左到右递增排序，且每行的第一个元素大于前一行的最后一个元素。文章提供了两种解决方案：一种是时间复杂度为O(log mn)的二分查找法；另一种是从右上角或左下角开始，逐步逼近目标值的方法，时间复杂度为O(m+n)。

题目描述

Write an efficient algorithm that searches for a value in anm x n matrix. This matrix has the following properties:

Integers in each row are sorted from left to right.
The first integer of each row is greater than the last integer of the previous row.

For example,

Consider the following matrix:

[
  [1,   3,  5,  7],
  [10, 11, 16, 20],
  [23, 30, 34, 50]
]

Given target =3, returntrue.

方法一：二分查找，效率最高，时间复杂度为O(logmn)

class Solution {
public:
    bool searchMatrix(vector<vector<int> > &matrix, int target) {
        int rows = matrix.size();
        int cols = matrix[0].size();
        int begin = 0,end = rows*cols-1;
         
        while(begin <= end)
        {
            int mid = (begin + end)/2;
            int row = mid/cols;
            int col = mid%cols;
            if(matrix[row][col] == target)
                return true;
            else if(matrix[row][col] < target)
                begin = mid + 1;
            else
                end = mid - 1;
        }
        return false;
    }
};

方法二：从右上角和左下角向中间靠拢，时间复杂度为O(m+n)

class Solution {
public:
bool searchMatrix(vector<vector<int> > &matrix, int target){
    int i=0, j=matrix[0].size()-1;
    while(i<matrix.size() && j>=0){
        if(matrix[i][j]==target){
            return true;
        }
        else if(matrix[i][j]<target){
            i++; //去掉这一行
        }
        else{
            j--;  //去掉这一列
        }
    }
    return false;
}
};