LeetCode 74. Search a 2D Matrix

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最新推荐文章于 2025-09-11 19:19:55 发布

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分类专栏：算法与OJ C/C++ 文章标签： leetcode

本文链接：https://blog.youkuaiyun.com/zhyh1435589631/article/details/51219877

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本文介绍了一种高效的矩阵搜索算法，用于在一特殊类型的二维矩阵中查找特定值。矩阵的每一行从左到右递增排序，且每行的第一个元素大于前一行的最后一个元素。文章提供了两种解题思路：逐行二分查找及将矩阵视为一维数组的整体二分查找，并附带了详细的C++代码实现。

1. 题目描述

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 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, return true.

2. 解题思路

拿到这个题目，很容易想到，这是行列都是有序的数组，那么我们就可以借鉴二分法的思想进行处理
先找到相应的行，然后找到值在该行中的位置

3. code

class Solution {
public:
    bool searchMatrix(vector<vector<int>>& matrix, int target) {
        int start = 0; int stop = matrix.size() - 1;
        int line = 0;
        // 查找所在行
        while (start < stop){
            int mid = start + ((stop - start) >> 1);
            if (matrix[mid][0] == target)
                return true;

            if (start == stop - 1){
                if (matrix[stop][0] > target)
                    line = start;
                else
                    line = stop;

                break;
            }

            if (matrix[mid][0] > target)
                stop = mid - 1;
            else
                start = mid;
        }

        // 不同的跳出方式
        if (start == stop)
            line = start;

        // 查找值
        start = 0, stop = matrix[0].size() - 1;
        while (start < stop){
            int mid = start + ((stop - start) >> 1);
            if (target == matrix[line][mid])
                return true;

            if (start == stop - 1){
                if (matrix[line][stop] == target)
                    return true;
                return false;
            }

            if (matrix[line][mid] > target)
                stop = mid - 1;
            else
                start = mid + 1;
        }

        return matrix[line][start] == target;
    }
};

4. 大神解法

这是一个整体有序的数组，可以整体看成是一个有序数组进行处理， brilliant！！！

/*
Use binary search.

n * m matrix convert to an array => matrix[x][y] => a[x * m + y]

an array convert to n * m matrix => a[x] =>matrix[x / m][x % m];
*/
class Solution {
public:
    bool searchMatrix(vector<vector<int> > &matrix, int target) {
        int n = matrix.size();
        int m = matrix[0].size();
        int l = 0, r = m * n - 1;
        while (l != r){
            int mid = (l + r - 1) >> 1;
            if (matrix[mid / m][mid % m] < target)
                l = mid + 1;
            else 
                r = mid;
        }
        return matrix[r / m][r % m] == target;
    }
};