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摘要: 1 public class Solution { 2 public int numSquares(int n) { 3 int[] dp = new int[n + 1]; 4 for (int i = 1; i <= n; i++) { 5 dp[i] = i; 6 for (int j = ... 阅读全文
posted @ 2016-07-07 13:11 keepshuatishuati 阅读(132) 评论(0) 推荐(0) 编辑
摘要: O (n) solution. 阅读全文
posted @ 2016-07-07 13:01 keepshuatishuati 阅读(143) 评论(0) 推荐(0) 编辑
摘要: This is almost the version of K list iterators. 阅读全文
posted @ 2016-07-07 12:56 keepshuatishuati 阅读(180) 评论(0) 推荐(0) 编辑
摘要: 1. Deal with 0 situation : only 0 itself is allowed, it cannot be 04. 2. We have to record the previous value that can be used to extract from previou 阅读全文
posted @ 2016-07-07 12:45 keepshuatishuati 阅读(118) 评论(0) 推荐(0) 编辑
摘要: 1 // Java Iterator interface reference: 2 // https://docs.oracle.com/javase/8/docs/api/java/util/Iterator.html 3 class PeekingIterator implements Iter 阅读全文
posted @ 2016-07-06 15:26 keepshuatishuati 阅读(120) 评论(0) 推荐(0) 编辑
摘要: Use level + 1 to avoid traversal two rooms back and forth. 阅读全文
posted @ 2016-07-06 15:17 keepshuatishuati 阅读(126) 评论(0) 推荐(0) 编辑
摘要: Remember: need to check whether has the word in hashset and garantee size is one. 阅读全文
posted @ 2016-07-06 14:33 keepshuatishuati 阅读(116) 评论(0) 推荐(0) 编辑
摘要: This is straight forward solution. 00 D <- D 10 L <- D 11 L <- L 01 D <- D So we need to differentiate 11 = 3 and 10 = 2 阅读全文
posted @ 2016-07-06 14:19 keepshuatishuati 阅读(156) 评论(0) 推荐(0) 编辑
摘要: 1 public class Solution { 2 public boolean canWin(String s) { 3 if (s.length() result = new ArrayList(); 7 for (int i = 0; i < s.length() - 1; i++) { 8 if (s.cha... 阅读全文
posted @ 2016-07-06 14:10 keepshuatishuati 阅读(109) 评论(0) 推荐(0) 编辑
摘要: 1 public class Solution { 2 public List generatePossibleNextMoves(String s) { 3 List result = new ArrayList(); 4 if (s.length() < 2) { 5 return result; 6 ... 阅读全文
posted @ 2016-07-06 09:39 keepshuatishuati 阅读(206) 评论(0) 推荐(0) 编辑
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