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最新评论
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likesky3:
看了数据结构书得知并不是迭代和递归的区别,yb君的写法的效果是 ...
Leetcode - Graph Valid Tree -
likesky3:
迭代和递归的区别吧~
Leetcode - Graph Valid Tree -
qb_2008:
还有一种find写法:int find(int p) { i ...
Leetcode - Graph Valid Tree -
qb_2008:
要看懂这些技巧的代码确实比较困难。我是这么看懂的:1. 明白这 ...
Leetcode - Single Num II -
qb_2008:
public int singleNumber2(int[] ...
Leetcode - Single Num II
Given a string S and a string T, count the number of distinct subsequences of T in S.
A subsequence of a string is a new string which is formed from the original string by deleting some (can be none) of the characters without disturbing the relative positions of the remaining characters. (ie, "ACE" is a subsequence of "ABCDE" while "AEC" is not).
Here is an example:
S = "rabbbit", T = "rabbit"
Return 3.
A subsequence of a string is a new string which is formed from the original string by deleting some (can be none) of the characters without disturbing the relative positions of the remaining characters. (ie, "ACE" is a subsequence of "ABCDE" while "AEC" is not).
Here is an example:
S = "rabbbit", T = "rabbit"
Return 3.
public class Solution { public int numDistinct(String s, String t) { if (s == null) return 0; if (t == null || t.length() == 0) return 1; if (s.length() == 0) return 0; int lenS = s.length(); int lenT = t.length(); int[][] dp = new int[2][lenT + 1]; dp[0][0] = 1; int lastLenS = 0; int currLenS = 1; for (int i = 1; i <= lenS; i++) { currLenS = 1 - lastLenS; dp[currLenS][0] = 1; for (int j = 1; j <= lenT; j++) { dp[currLenS][j] = dp[lastLenS][j]; if (s.charAt(i - 1) == t.charAt(j - 1)) dp[currLenS][j] += dp[lastLenS][j-1]; } lastLenS = currLenS; } return dp[currLenS][lenT]; } // dp[i][j] = dp[i - 1][j] + (s[i - 1] == t[j - 1] ? dp[i - 1][j - 1] : 0); // dp[i][j]表示s的前i字符构成的子串能匹配t中前j个字符构成的子串的子序列数 // 不管s和t当前考察位置处的字符是否相等,dp[i][j]至少有dp[i-1][j] // 如果当前考察位置处两字符相等,则再加上dp[i-1][j-1]的数目 public int numDistinct1(String s, String t) { if (s == null) return 0; if (t == null || t.length() == 0) return 1; if (s.length() == 0) return 0; int lenS = s.length(); int lenT = t.length(); int[][] dp = new int[lenS + 1][lenT + 1]; dp[0][0] = 1; for (int i = 1; i <= lenS; i++) { dp[i][0] = 1; for (int j = 1; j <= lenT; j++) { dp[i][j] = dp[i - 1][j]; if (s.charAt(i - 1) == t.charAt(j - 1)) dp[i][j] += dp[i - 1][j - 1]; } } return dp[lenS][lenT]; } }
发表评论
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Leetcode - Ugly Number II
2015-08-24 22:54 1127[分析] 暴力的办法就是从1开始检查每个数是否是丑数,发现丑数 ... -
Leetcode - Paint House II
2015-08-20 20:37 1577There are a row of n houses, ea ... -
Leetcode - Maximum Square
2015-08-16 13:33 472Given a 2D binary matrix filled ... -
Leetcode - Paint House
2015-08-16 10:48 1117There are a row of n houses, ea ... -
Leetcode - Different Ways to Add Parentheses
2015-07-29 20:21 1157Given a string of numbers and o ... -
Jump Game II
2015-07-05 16:49 504Given an array of non-negative ... -
Leetcode - Jump Game
2015-07-05 15:52 491Given an array of non-negative ... -
Leetcode - Interleaving String
2015-06-07 11:41 569Given s1, s2, s3, find whe ... -
Leetcode - Wildcard Matching
2015-06-06 20:01 954Implement wildcard pattern ma ... -
Leetcode - Maximal Square
2015-06-04 08:25 592Given a 2D binary matrix fille ... -
Leetcode - Palindrome Partition II
2015-05-21 21:15 646Given a string s, partition ... -
Leetcode - Palindrome Partition
2015-05-21 09:56 751Given a string s, partition s s ... -
Leetcode - House Robber II
2015-05-20 22:34 734Note: This is an extension of ... -
Leetcode - Maximum Rectangle
2015-05-20 08:58 466Given a 2D binary matrix fill ... -
Leetcode - Scramble String
2015-05-17 14:22 547Given a string s1, we may repre ... -
Leetcode - Regular Expression Matching
2015-05-16 16:31 387Implement regular expression ma ... -
Leetcode - Best Time to Buy and Sell Stock IV
2015-05-01 16:11 584Say you have an array for which ... -
Leetcode - Best Time to Buy and Sell Stock IV
2015-04-23 09:59 0public class Solution ... -
Leetcode - Best Time to Buy and Sell Stock III
2015-04-23 09:04 452Say you have an array for whi ... -
Leetcode - Dungeon Game
2015-04-21 09:50 419The demons had captured the pr ...
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