2019年3月28日
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Another Two Sum java class Solution { public int subarraysDivByK(int[] A, int K) { int[] sum = new int[A.length + 1]; for (int i = 0; i c = new HashMa 阅读全文
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technically, no different with two sum 阅读全文
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This remains me of some 'subarray count' type problems….. 阅读全文
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Prefix product or two pointer can both solve this problem. But don't forget to take a look at the note 阅读全文
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prefix sum k = 0 is a corner case, in which case just save , not And every time, record the leftmost index of a key.(greedy) 阅读全文
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Use prefix product(sum? actually much the same). 阅读全文
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using sliding window And try to Using prefix sum method, the rightmost index with value = s ? 1: N + 1; for (int i = 1; i target) { ret = Math.min(ret 阅读全文
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O(N) space and O(M N) time solution, where M, N is A's row number and column number java class Solution { public int minFallingPathSum(int[][] A) { in 阅读全文
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Divide and conquer I'm not so good at this kind of problem, so I just checked the solution page. 阅读全文
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```java class RecentCounter { Queue q; public RecentCounter() { q = new LinkedList(); } public int ping(int t) { int threshold = t 3000; while (q.size 阅读全文