C++实现LeetCode(115.不同的子序列)

[LeetCode] 115. Distinct Subsequences 不同的子序列

Given a string S and a string T, count the number of distinct subsequences of S which equals T.

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).

Example 1:

Input: S =

"rabbbit"

, T =

"rabbit"
Output: 3

Explanation:

As shown below, there are 3 ways you can generate "rabbit" from S.
(The caret symbol ^ means the chosen letters)

rabbbit

^^^^ ^^

rabbbit

^^ ^^^^

rabbbit

^^^ ^^^

Example 2:

Input: S =

"babgbag"

, T =

"bag"
Output: 5

Explanation:

As shown below, there are 5 ways you can generate "bag" from S.
(The caret symbol ^ means the chosen letters)

babgbag

^^ ^

babgbag

^^    ^

babgbag

^    ^^

babgbag

  ^  ^^

babgbag

    ^^^

看到有关字符串的子序列或者配准类的问题，首先应该考虑的就是用动态规划 Dynamic Programming 来求解，这个应成为条件反射。而所有 DP 问题的核心就是找出状态转移方程，想这道题就是递推一个二维的 dp 数组，其中 dp[i][j] 表示s中范围是 [0, i] 的子串中能组成t中范围是 [0, j] 的子串的子序列的个数。下面我们从题目中给的例子来分析，这个二维 dp 数组应为：

  Ø r a b b b i t
Ø 1 1 1 1 1 1 1 1
r 0 1 1 1 1 1 1 1
a 0 0 1 1 1 1 1 1
b 0 0 0 1 2 3 3 3
b 0 0 0 0 1 3 3 3
i 0 0 0 0 0 0 3 3
t 0 0 0 0 0 0 0 3

首先，若原字符串和子序列都为空时，返回1，因为空串也是空串的一个子序列。若原字符串不为空，而子序列为空，也返回1，因为空串也是任意字符串的一个子序列。而当原字符串为空，子序列不为空时，返回0，因为非空字符串不能当空字符串的子序列。理清这些，二维数组 dp 的边缘便可以初始化了，下面只要找出状态转移方程，就可以更新整个 dp 数组了。我们通过观察上面的二维数组可以发现，当更新到 dp[i][j] 时，dp[i][j] >= dp[i][j - 1] 总是成立，再进一步观察发现，当 T[i - 1] == S[j - 1] 时，dp[i][j] = dp[i][j - 1] + dp[i - 1][j - 1]，若不等， dp[i][j] = dp[i][j - 1]，所以，综合以上，递推式为：

dp[i][j] = dp[i][j - 1] + (T[i - 1] == S[j - 1] ? dp[i - 1][j - 1] : 0)

根据以上分析，可以写出代码如下：

class Solution {
public:
    int numDistinct(string s, string t) {
        int m = s.size(), n = t.size();
        vector<vector<long>> dp(n + 1, vector<long>(m + 1));
        for (int j = 0; j <= m; ++j) dp[0][j] = 1;
        for (int i = 1; i <= n; ++i) {
            for (int j = 1; j <= m; ++j) {
                dp[i][j] = dp[i][j - 1] + (t[i - 1] == s[j - 1] ? dp[i - 1][j - 1] : 0);
            }
        }
        return dp[n][m];
    }
};

Github 同步地址：

https://github.com/grandyang/leetcode/issues/115

参考资料：

https://leetcode.com/problems/distinct-subsequences/

https://leetcode.com/problems/distinct-subsequences/discuss/37327/Easy-to-understand-DP-in-Java

https://leetcode.com/problems/distinct-subsequences/discuss/37412/Any-better-solution-that-takes-less-than-O(n2)-space-while-in-O(n2)-time

到此这篇关于C++实现LeetCode(115.不同的子序列)的文章就介绍到这了,更多相关C++实现不同的子序列内容请搜索我们以前的文章或继续浏览下面的相关文章希望大家以后多多支持我们！

赞 (0)

相关推荐

  • C++实现LeetCode(141.单链表中的环)

    C++实现LeetCode(141.单链表中的环)

    [LeetCode] 141. Linked List Cycle 单链表中的环 Given a linked list, determine if it has a cycle in it. To represent a cycle in the given linked list, we use an integer pos which represents the position (0-indexed) in the linked list where tail connects to.

  • C++实现LeetCode(889.由先序和后序遍历建立二叉树)

    [LeetCode] 889. Construct Binary Tree from Preorder and Postorder Traversal 由先序和后序遍历建立二叉树 Return any binary tree that matches the given preorder and postorder traversals. Values in the traversals pre and post are distinct positive integers. Example 1

  • C++实现LeetCode(105.由先序和中序遍历建立二叉树)

    [LeetCode] 105. Construct Binary Tree from Preorder and Inorder Traversal 由先序和中序遍历建立二叉树 Given preorder and inorder traversal of a tree, construct the binary tree. Note: You may assume that duplicates do not exist in the tree. For example, given preor

  • C++实现LeetCode(142.单链表中的环之二)

    C++实现LeetCode(142.单链表中的环之二)

    [LeetCode] 142. Linked List Cycle II 单链表中的环之二 Given a linked list, return the node where the cycle begins. If there is no cycle, return null. To represent a cycle in the given linked list, we use an integer pos which represents the position (0-indexe

  • C++实现LeetCode(106.由中序和后序遍历建立二叉树)

    [LeetCode] 106. Construct Binary Tree from Inorder and Postorder Traversal 由中序和后序遍历建立二叉树 Given inorder and postorder traversal of a tree, construct the binary tree. Note: You may assume that duplicates do not exist in the tree. For example, given ino

  • C++实现LeetCode(109.将有序链表转为二叉搜索树)

    [LeetCode] 109.Convert Sorted List to Binary Search Tree 将有序链表转为二叉搜索树 Given a singly linked list where elements are sorted in ascending order, convert it to a height balanced BST. For this problem, a height-balanced binary tree is defined as a binary

  • C++实现LeetCode(110.平衡二叉树)

    [LeetCode] 110.Balanced Binary Tree 平衡二叉树 Given a binary tree, determine if it is height-balanced. For this problem, a height-balanced binary tree is defined as: a binary tree in which the depth of the two subtrees of everynode never differ by more t

  • C++实现LeetCode(143.链表重排序)

    [LeetCode] 143.Reorder List 链表重排序 Given a singly linked list L: L0→L1→-→Ln-1→Ln, reorder it to: L0→Ln→L1→Ln-1→L2→Ln-2→- You may not modify the values in the list's nodes, only nodes itself may be changed. Example 1: Given 1->2->3->4, reorder it t

  • C++实现LeetCode(115.不同的子序列)

    [LeetCode] 115. Distinct Subsequences 不同的子序列 Given a string S and a string T, count the number of distinct subsequences of S which equals T. A subsequence of a string is a new string which is formed from the original string by deleting some (can be n

  • C++实现leetcode(3.最长无重复字符的子串)

    [LeetCode] 3. Longest Substring Without Repeating Characters 最长无重复字符的子串 Given a string, find the length of the longest substring without repeating characters. Example 1: Input: "abcabcbb" Output: 3 Explanation: The answer is "abc", wit

  • C++实现LeetCode(76.最小窗口子串)

    [LeetCode] 76. Minimum Window Substring 最小窗口子串 Given a string S and a string T, find the minimum window in S which will contain all the characters in T in complexity O(n). Example: Input: S = "ADOBECODEBANC", T = "ABC" Output: "BA

  • C++实现LeetCode(93.复原IP地址)

    [LeetCode] 93.Restore IP Addresses 复原IP地址 Given a string containing only digits, restore it by returning all possible valid IP address combinations. Example: Input: "25525511135" Output: ["255.255.11.135", "255.255.111.35"] 这

  • C++实现LeetCode(97.交织相错的字符串)

    [LeetCode] 97.Interleaving String 交织相错的字符串 Given s1, s2, s3, find whether s3 is formed by the interleaving of s1 and s2. Example 1: Input: s1 = "aabcc", s2 = "dbbca", s3 = "aadbbcbcac" Output: true Example 2: Input: s1 = &quo

  • C++ LeetCode300最长递增子序列

    目录 LeetCode 300.最长递增子序列 方法一:动态规划 AC代码 C++ LeetCode 300.最长递增子序列 力扣题目链接:leetcode.cn/problems/lo… 给你一个整数数组 nums ,找到其中最长严格递增子序列的长度. 子序列 是由数组派生而来的序列,删除(或不删除)数组中的元素而不改变其余元素的顺序.例如,[3,6,2,7] 是数组 [0,3,1,6,2,2,7] 的子序列. 示例 1: 输入:nums = [10,9,2,5,3,7,101,18]输出:4

  • PHP求最大子序列和的算法实现

    复制代码 代码如下: <?php //作者:遥远的期待 //QQ:15624575 //算法分析:1.必须是整数序列.2.如果整个序列不全是负数,最大子序列的第一项必须是正数,否则最大子序列后面的数加起来再加上第一项的负数,其和肯定不是最大的:3.如果整个序列都是负数,那么最大子序列的和是0: //全负数序列很简单,不举例 $arr=array(4,-3,5,-2,-1,2,6,-2); function getmaxsum($arr){ $thissum=0; $maxsum=0; $star

  • LeetCode -- Path Sum III分析及实现方法

    LeetCode -- Path Sum III分析及实现方法 题目描述: You are given a binary tree in which each node contains an integer value. Find the number of paths that sum to a given value. The path does not need to start or end at the root or a leaf, but it must go downwards

  • SQL计算字符串中最大的递增子序列的方法

    SQL计算字符串中最大的递增子序列的方法

    求字符串中最大的递增子序列 数据库环境:SQL SERVER 2005 如题,求字符串"abcbklmnodfghijkmer"中最大的递增子序列.这个字符串有点特别, 只由26个小写字母a-z组成. 大概思路如下: 1.将字符串转到一列存储,并生成行号 2.设置一个递增计数器列,默认为1,比较上下行的字符,如果在字典中的顺序是递增, 则计数器加1,否则,计数器置1 3.找出计数器最大的数及对应的行号,根据这2个数截取字符串 思路有了,下面直接贴代码 DECLARE @vtext VA

  • LIS 最长递增子序列 Java的简单实现

    今天遇到了一个求最长递增子序列的问题,看了之后就尝试着用Java实现了一下,关于什么是最长递增子序列,这里就不在赘述,可以百度或者Google之,以下为实现的代码: 说明:本段代码实现的功能为 (1)随机生成一个有10个元素的数组,然后输出它的最长递增子序列 (2)输出以其中某一个元素为结尾的最长递增子序列的长度 具体的实现思路在注释中已经详细表明了,比较简单,这里就不再赘述 import java.util.Arrays; import java.util.Random; public cla

随机推荐