05 August 2008
给两个整数数组，找出共有的最长的连续子数组的长度
DP, dp[i][j] is the length of longest common subarray ending with nums[i-1] and nums[j-1]
dp[i][j] = dp[i - 1][j - 1] + 1;
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class Solution {
public int findLength(int[] nums1, int[] nums2) {
int len1 = nums1.length;
int len2 = nums2.length;
int result = 0;
int[][] dp = new int[len1 + 1][len2 + 1];
for (int i = 1; i < len1 + 1; i++) {
for (int j = 1; j < len2 + 1; j++) {
if (nums1[i - 1] == nums2[j - 1]) {
dp[i][j] = dp[i - 1][j - 1] + 1;
result = Math.max(result, dp[i][j]);
}
}
}
return result;
}
}
滑动窗口也可以
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class Solution {
public int findLength(int[] A, int[] B) {
return A.length < B.length ? findMax(A, B) : findMax(B, A);
}
int findMax(int[] A, int[] B) {
int max = 0;
int an = A.length, bn = B.length;
for (int len = 1; len <= an; len++) {
max = Math.max(max, maxLen(A, 0, B, bn - len, len));
}
for (int j = bn - an; j >= 0; j--) {
max = Math.max(max, maxLen(A, 0, B, j, an));
}
for (int i = 1; i < an; i++) {
max = Math.max(max, maxLen(A, i, B, 0, an - i));
}
return max;
}
int maxLen(int[] a, int i, int[] b, int j, int len) {
int count = 0, max = 0;
for (int k = 0; k < len; k++) {
if (a[i + k] == b[j + k]) {
count++;
} else if (count > 0) {
max = Math.max(max, count);
count = 0;
}
}
return count > 0 ? Math.max(max, count) : max;
}
}