[LeetCode] Is Subsequence 题解

前言

这道题的实现方法有很多，包括dp，贪心算法，二分搜索，普通实现等等。

题目

Given a string s and a string t, check if s is subsequence of t.

You may assume that there is only lower case English letters in both s and t. t is potentially a very long (length ~= 500,000) string, and s is a short string (<=100).

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:

s = "abc", t = "ahbgdc"

Return true.

Example 2:

s = "axc", t = "ahbgdc"

Return false.

Follow up:

If there are lots of incoming S, say S1, S2, ... , Sk where k >= 1B, and you want to check one by one to see if T has its subsequence. In this scenario, how would you change your code?

题意说的是，判断字符串s是否是字符串t的字串（要求按照s中字符的顺序！）

实现

//
//  Is Subsequence.cpp
//  LeetCodeCppPro
//
//  Created by George on 17/3/2.
//  Copyright © 2017年 George. All rights reserved.
//

#include <stdio.h>
#include "PreLoad.h"

class Solution {
public:
    /**
     *  两个节点，贪心算法的思想
     *
     *  @param s <#s description#>
     *  @param t <#t description#>
     *
     *  @return <#return value description#>
     */
    bool isSubsequence(string s, string t) {
        if (s == t || (s == "" && t.length() > s.length())) {
            return true;
        }
        else if (s.length() > t.length()) {
            return false;
        }
        int idx = 0;
        for (int  i = 0; i < s.length(); i++,idx++) {
            while (idx < t.length() && t[idx] != s[i]) {
                idx++;
            }

            if (idx >= t.length()) {
                return false;
            }
        }

        return true;
    }

    /**
     *  two point
     *  失败，没调试好
     *
     *  @param s <#s description#>
     *  @param t <#t description#>
     *
     *  @return <#return value description#>
     */
    bool isSubsequence2(string s, string t) {
        if (s == t || (s == "" && t.length() > s.length())) {
            return true;
        }
        else if (s.length() > t.length() || (s.length() == t.length() && s != t)) {
            return false;
        }

        int s_left = 0, s_right = s.length()-1;
        int t_left = 0, t_right = t.length()-1;

        while (t_left < t_right) {
            while (s[s_left] != t[t_left] && t_left < t_right) {
                t_left++;
            }
            while (s[s_right] != t[t_right] && t_left < t_right) {
                t_right--;
            }

            if (t_left == t_right) {
                if (s[s_left] == t[t_left]) {
                    return true;
                }
                else {
                    return false;
                }
            }

            if (s[s_left] == t[t_left]) {
                s_left++;
                t_left++;
            }
            else {
                t_left++;
            }

            if (s[s_right] == t[s_right]) {
                s_right--;
                t_right--;
            }
            else {
                t_right--;
            }

            if (t_left == t_right) {
                if (s[s_left] == t[t_left]) {
                    return true;
                }
                else {
                    return false;
                }
            }
        }

        if (t_left >= t_right) {
            return false;
        }
        else {
            return true;
        }
    }

    /**
     *  使用hash，取idx递增顺序
     *
     *  @param s <#s description#>
     *  @param t <#t description#>
     *
     *  @return <#return value description#>
     */
    bool isSubsequence3(string s, string t) {
        vector<vector<int>> sequence(26);

        // 纪录下每个单词出现的下标
        for (int i = 0; i < t.length(); i++) {
            int idx = t[i] - 'a';
            sequence[idx].push_back(i);
        }

        // 根据s的单词顺序
        int preIndex = -1;
        for (int i = 0; i < s.length(); i++) {
            int idx = s[i] - 'a';
            auto nums = sequence[idx]; //取得字符的下标集合
            if (nums.empty()) {
                return false;
            }

            auto itr2 = nums.begin();
            while (itr2 != nums.end() && *itr2 <= preIndex) {
                itr2++;
            }

            if (itr2 == nums.end()) {
                return false;
            }
            else {
                preIndex = *itr2;
            }
        }

        return true;
    }

    /**
     *  和上面的思想类似，
     *  不同的是使用了一个数组index记录下同一个单词的次数，读取更加方便
     *
     *  @param s <#s description#>
     *  @param t <#t description#>
     *
     *  @return <#return value description#>
     */
    bool isSubsequence4(string s, string t) {
        vector<vector<int>> posMap(26); // 题目已经假设是小写字母

        // 纪录目标字符串t中字符的出现次数
        for (int i = 0; i < t.length(); i++) {
            posMap[t[i] - 'a'].push_back(i);
        }

        int pre = -1;
        int index[26];  //纪录s中每个单词出现的个数，为后面取该字符的下标做准备，避免从头开始遍历
        memset(index, -1, sizeof(index));
        for (int i = 0; i < s.length(); i++) {
            int j = s[i] - 'a';
            index[j]++;

            // posMap[j][index[j]] 为取某个字符的下标
            while (index[j] < posMap[j].size()) {
                if (posMap[j][index[j]] > pre) {  //直到取得比上一个字符的下标大的下标值
                    break;
                }
                ++index[j];
            }

            if (index[j] > posMap[j].size()) {
                return false;
            }

            pre = posMap[j][index[j]]; //更新为当前字符的下标
        }

        return true;
    }

    /**
     *  思路和前面的一样，都使用了之前的下标
     *  不同的是做法更佳简洁，因为使用了upper_bound函数
     *
     *  @param s <#s description#>
     *  @param t <#t description#>
     *
     *  @return <#return value description#>
     */
    bool isSubsequence5(string s, string t) {
        vector<vector<int>> record(26);

        for (int i = 0; i < t.size(); i++) {
            record[t[i] - 'a'].push_back(i);
        }

        int index = -1;
        for (int i = 0; i < s.size(); i++) {
            int idx = s[i] - 'a';
            auto itr = upper_bound(record[idx].begin(), record[idx].end(), index);
            if (itr == record[idx].end()) {
                return false;
            }
            index = *itr;
        }

        return true;
    }

    /**
     *  和第一种思想类似
     *
     *  @param s <#s description#>
     *  @param t <#t description#>
     *
     *  @return <#return value description#>
     */
    bool isSubsequence6(string s, string t) {
        if (s.length() == 0) {
            return true;
        }
        queue<int> queue; //使用队列可以保证字符的处理顺序是正确的
        for (char c : s) queue.push(c);
        for (int i = 0; !queue.empty() && i < t.length(); i++) {
            if (t[i] == queue.front()) {
                queue.pop();
            }
        }
        return queue.empty();
    }

    /**
     *  dp实现
     *  不同点在于，只使用了两行数据，节省空间
     *  使用了之后到的数据覆盖之前的数据
     *
     *  @param s <#s description#>
     *  @param t <#t description#>
     *
     *  @return <#return value description#>
     */
    bool isSubsequence7(string s, string t) {
        if (s.length() == 0) {
            return true;
        }
        else if (t.length() != 0) {
            return false;
        }

        int m = t.length(), n = s.length();
        vector<vector<bool>> dp(2, vector<bool>(n+1, false));

        // 设置第一行，因为只需要看s的数据
        for (int i = 0; i <= n; i++) {
            dp[0][i] = true;
        }

        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (t[i] == s[j]) {
                    dp[1][j+1] = dp[0][j];
                }
                else {
                    dp[1][j+1] = dp[1][j];
                }
            }

            // 进行覆盖，将第1行数据覆盖到第0层数据上，循环使用
            for (int j = 0; j < n; j++) {
                dp[0][j] = dp[1][j];
            }
        }

        return dp[1][n];
    }

    /**
     *  正儿八经的使用dp
     *
     *  @param s <#s description#>
     *  @param t <#t description#>
     *
     *  @return <#return value description#>
     */
    bool isSubsequence8(string s, string t) {
        if (s.length() == 0) {
            return true;
        }
        else if (t.length() != 0) {
            return false;
        }

        int m = t.length(), n = s.length();
        vector<vector<bool>> dp(m, vector<bool>(n, false));

        if (s[0] == t[0]) {
            dp[0][0] = true;
        }

        // 将第一列的数据置为false，因为不需要
        for (int i = 1; i  < m; i++) {
            dp[i][0] = false;
        }

        // 第一行数据
        for (int i = 1; i < n; i++) {
            dp[0][i] = dp[0][i-1] || s[i] == t[0];
        }

        for (int i = 1; i < m; i++) {
            for (int j = 1; j < n; j++) {
                if (t[i] == s[j]) {
                    dp[i][j] = dp[i-1][j-1] || dp[i][j-1];
                    // 将该行上的数据都置为true
                    if (dp[i][j]) {
                        for (int k = j+1; k < n; k++) {
                            dp[i][k] = true;
                        }
                        continue;
                    }
                }
            }
        }

        return dp[m-1][n-1];
    }

    void test() {
        string s = "aaaabc", t = "llallllbllllc";

        if (isSubsequence3(s, t)) {
            cout << "true" << endl;
        }
        else {
            cout << "false" << endl;
        }
    }
};