Leetcode:718. Maximum Length of Repeated Subarray

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本文介绍了一种求解两个整数数组中最长公共子数组的方法。通过两种途径实现：一是遍历匹配方法，适用于简单直观理解；二是采用动态规划方式，更高效地解决了重复子问题，提高了算法效率。

Description

Given two integer arrays A and B, return the maximum length of an subarray that appears in both arrays.

Example 1:
Input:
A: [1,2,3,2,1]
B: [3,2,1,4,7]
Output: 3
Explanation:
The repeated subarray with maximum length is [3, 2, 1].
Note:
1 <= len(A), len(B) <= 1000
0 <= A[i], B[i] < 100

解题思路

本题求最长公共子数组问题，可采用最容易想到的遍历匹配方法求解

int findLength(vector<int>& A, vector<int>& B) {
        int max = 0;
        int count = 0;
        int j = 0;
        int lengthA = A.size();
        int lengthB = B.size();
        for (int i = 0; i < lengthA; ++i) {
            count = 0;
            j = 0;
            if (lengthA - i <= max) break;
            while (j != lengthB) {
                if (i + count < lengthA && j + count < lengthB && A[i + count] == B[j + count]) {
                    ++count;
                } else {
                    if (max < count) {
                        max = count;
                    }
                    count = 0;
                    ++j;
                }
            }
        }
        return max;
    }

本题中有很多重复子问题，也可用动态规划进行求解
动态方程为
dp[i][j] = dp[i - 1][j - 1] + 1;
返回结果为
max = dp[i][j] > max ? dp[i][j] : max;

class Solution {
public:
    int findLength(vector<int>& A, vector<int>& B) {
        int len1 = A.size();
        int len2 = B.size();
        int max = 0;
        //初始化赋值为0
        vector<vector<int>> dp(len1+1, vector<int>(len2+1,0));
        for(int i = 1; i <= len1; i++) {
            for (int j = 1; j <= len2; j++) {
                if (A[i-1] == B[j-1]) {
                    dp[i][j] = dp[i-1][j-1]+1;
                    max = dp[i][j] > max ? dp[i][j] :max;
                }
            }
        }
        return max;

    }
};