Given two integer arrays nums1 and nums2, return the maximum length of a subarray that appears in both arrays.
Example 1:
Input: nums1 = [1,2,3,2,1], nums2 = [3,2,1,4,7] Output: 3 Explanation: The repeated subarray with maximum length is [3,2,1].
Example 2:
Input: nums1 = [0,0,0,0,0], nums2 = [0,0,0,0,0] Output: 5
Constraints:
1 <= nums1.length, nums2.length <= 10000 <= nums1[i], nums2[i] <= 100class Solution {
public:
int findLength(vector<int>& nums1, vector<int>& nums2) {
int m = nums1.size(), n = nums2.size();
if (m < n) return findLength(nums2, nums1);
vector<int> dp(n+1), prevDP(n+1);
int ans = 0;
for (int i = 1; i <= m; ++i) {
for (int j = 1; j <= n; ++j) {
if (nums1[i-1] == nums2[j-1])
dp[j] = prevDP[j-1] + 1;
else dp[j] = 0;
ans = max(ans, dp[j]);
}
dp.swap(prevDP);
}
return ans;
}
};