https://leetcode.com/problems/arithmetic-slices-ii-subsequence/

太难了。。。

package com.company;

import java.util.*;

// 终于找到原因了。还是我重复加了
// 比如 46 46 47 48 49 // 用累积记录过往节点记录,发现超时了
// 参考了Discuss
// https://discuss.leetcode.com/topic/66725/o-n-2-mle-tle-in-c-try-this-one-concise-and-fast // 其中 +1 这个地方 想了很久,终于想通了。意味着之前有一个长度为2的,会变成3.而为什么只加1就可以呢,因为用的是 last but one,
// 倒数第二个,帮忙记录最后一个可能的结果,而倒数第二个和前一个的连接,总是一对一的。。 // 实在太难了 class Solution {
public int numberOfArithmeticSlices(int[] A) {
// Need use Long to avoid gap exceeding
Map<Integer, Map<Long, Integer>> mp = new HashMap<>();
Set<Long> cSet = new HashSet<>();
for (int k: A) {
cSet.add((long)k);
} int ret = 0;
Map<Long, Integer> lastMp;
long gap;
int tmp;
int curRet; for (int i=0; i<A.length; i++) {
Map<Long, Integer> tmpMp = new HashMap<>(); for (int j=i-1; j>=0; j--) { gap = (long)A[i] - (long)A[j]; if (tmpMp.containsKey(gap)) {
curRet = tmpMp.get(gap);
}
else {
curRet = 0;
} lastMp = mp.get(j);
if (!lastMp.containsKey(gap)) {
tmp = 0;
}
else {
tmp = lastMp.get(gap);
ret += tmp;
} //System.out.printf("val %d gap %d\n", A[i], gap); if (cSet.contains(A[i] + gap)) {
tmpMp.put(gap, curRet + tmp + 1);
//System.out.printf("here val %d gap %d\n", A[i], gap);
}
} mp.put(i, tmpMp);
}
return ret;
}
} public class Main { public static void main(String[] args) throws InterruptedException { System.out.println("Hello!");
Solution solution = new Solution(); // Your Codec object will be instantiated and called as such:
int[] A = {1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1};
int ret = solution.numberOfArithmeticSlices(A);
System.out.printf("ret:%d\n", ret); System.out.println(); } }

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