class Solution {
public:
int findLength(vector<int>& nums1, vector<int>& nums2) {
}
};
718. 最长重复子数组
给两个整数数组 nums1 和 nums2 ，返回 两个数组中 公共的 、长度最长的子数组的长度 。
示例 1：
输入：nums1 = [1,2,3,2,1], nums2 = [3,2,1,4,7] 输出：3 解释：长度最长的公共子数组是 [3,2,1] 。
示例 2：
输入：nums1 = [0,0,0,0,0], nums2 = [0,0,0,0,0] 输出：5
提示：
1 <= nums1.length, nums2.length <= 10000 <= nums1[i], nums2[i] <= 100相似题目
原站题解
python3 解法, 执行用时: 1940 ms, 内存消耗: 14.9 MB, 提交时间: 2022-12-06 13:35:57
class Solution:
def findLength(self, A: List[int], B: List[int]) -> int:
def maxLength(addA: int, addB: int, length: int) -> int:
ret = k = 0
for i in range(length):
if A[addA + i] == B[addB + i]:
k += 1
ret = max(ret, k)
else:
k = 0
return ret
n, m = len(A), len(B)
ret = 0
for i in range(n):
length = min(m, n - i)
ret = max(ret, maxLength(i, 0, length))
for i in range(m):
length = min(n, m - i)
ret = max(ret, maxLength(0, i, length))
return ret
golang 解法, 执行用时: 20 ms, 内存消耗: 2.9 MB, 提交时间: 2022-12-06 13:35:44
func findLength(A []int, B []int) int {
n, m := len(A), len(B)
ret := 0
for i := 0; i < n; i++ {
len := min(m, n - i)
maxLen := maxLength(A, B, i, 0, len)
ret = max(ret, maxLen)
}
for i := 0; i < m; i++ {
len := min(n, m - i)
maxLen := maxLength(A, B, 0, i, len)
ret = max(ret, maxLen)
}
return ret
}
func maxLength(A, B []int, addA, addB, len int) int {
ret, k := 0, 0
for i := 0; i < len; i++ {
if A[addA + i] == B[addB + i] {
k++
} else {
k = 0
}
ret = max(ret, k)
}
return ret
}
func max(x, y int) int {
if x > y {
return x
}
return y
}
func min(x, y int) int {
if x < y {
return x
}
return y
}
java 解法, 执行用时: 28 ms, 内存消耗: 41.3 MB, 提交时间: 2022-12-06 13:35:31
class Solution {
public int findLength(int[] A, int[] B) {
int n = A.length, m = B.length;
int ret = 0;
for (int i = 0; i < n; i++) {
int len = Math.min(m, n - i);
int maxlen = maxLength(A, B, i, 0, len);
ret = Math.max(ret, maxlen);
}
for (int i = 0; i < m; i++) {
int len = Math.min(n, m - i);
int maxlen = maxLength(A, B, 0, i, len);
ret = Math.max(ret, maxlen);
}
return ret;
}
public int maxLength(int[] A, int[] B, int addA, int addB, int len) {
int ret = 0, k = 0;
for (int i = 0; i < len; i++) {
if (A[addA + i] == B[addB + i]) {
k++;
} else {
k = 0;
}
ret = Math.max(ret, k);
}
return ret;
}
}
java 解法, 执行用时: 24 ms, 内存消耗: 49.9 MB, 提交时间: 2022-12-06 13:35:13
class Solution {
public int findLength(int[] A, int[] B) {
int n = A.length, m = B.length;
int[][] dp = new int[n + 1][m + 1];
int ans = 0;
for (int i = n - 1; i >= 0; i--) {
for (int j = m - 1; j >= 0; j--) {
dp[i][j] = A[i] == B[j] ? dp[i + 1][j + 1] + 1 : 0;
ans = Math.max(ans, dp[i][j]);
}
}
return ans;
}
}
golang 解法, 执行用时: 40 ms, 内存消耗: 16 MB, 提交时间: 2022-12-06 13:34:57
func findLength(A []int, B []int) int {
n, m := len(A), len(B)
dp := make([][]int, n + 1)
for i := 0; i < len(dp); i++ {
dp[i] = make([]int, m + 1)
}
ans := 0
for i := n - 1; i >= 0; i-- {
for j := m - 1; j >= 0; j-- {
if A[i] == B[j] {
dp[i][j] = dp[i + 1][j + 1] + 1
} else {
dp[i][j] = 0
}
if ans < dp[i][j] {
ans = dp[i][j]
}
}
}
return ans
}
python3 解法, 执行用时: 3356 ms, 内存消耗: 39.6 MB, 提交时间: 2022-12-06 13:34:43
'''
令 dp[i][j] 表示 A[i:] 和 B[j:] 的最长公共前缀，那么答案即为所有 dp[i][j] 中的最大值
'''
class Solution:
def findLength(self, A: List[int], B: List[int]) -> int:
n, m = len(A), len(B)
dp = [[0] * (m + 1) for _ in range(n + 1)]
ans = 0
for i in range(n - 1, -1, -1):
for j in range(m - 1, -1, -1):
dp[i][j] = dp[i + 1][j + 1] + 1 if A[i] == B[j] else 0
ans = max(ans, dp[i][j])
return ans