LC 718. Maximum Length of Repeated Subarray
阿新 • • 發佈:2019-01-13
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
Runtime: 78 ms, faster than 40.98% of Java online submissions for Maximum Length of Repeated Subarray.
class Solution { public int findLength(int[] A, int[] B) { int[][] dp = new int[A.length+1][B.length+1]; intret = 0; for(int i=1; i<dp.length; i++){ for(int j=1; j<dp[0].length; j++){ if(A[i-1] == B[j-1]){ dp[i][j] = dp[i-1][j-1] + 1; ret = Math.max(ret, dp[i][j]); } } } return ret; } }
a better solution
Runtime: 26 ms, faster than 99.59% of Java online submissions for Maximum Length of Repeated Subarray.
有兩個關鍵點,
第一個是內層迴圈從後往前遍歷,如果從前往後遍歷就是錯誤的,因為我們每一次更新dp的時候是dp[j+1] = dp[j] + 1,所以在更新j+1的時候要用到j的資訊,而這個j應該是之前的j,也就是上一行的j,可以參考上面一個解法中矩陣的上一行。
第二個是如果沒有匹配到,應該把j+1變成0,否則會產生錯誤的計數。
class Solution { public int findLength(int[] A, int[] B) { int[] dp = new int[A.length+1]; int max = 0; for(int i=0; i<A.length; i++) { for(int j=B.length-1; j>=0; j--) { if (A[i] == B[j]) { dp[j+1] = dp[j] + 1; if (max < dp[j+1]) { max = dp[j+1]; } } else { dp[j+1] = 0; } } } return max; } }