(算法分析Week13)Is Subsequence[Medium]

506. Relative Ranks[Easy]

题目来源

Description

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.

Solution

动态规划分类下的题目，其实没感觉典型动态规划。
就两个指针往后扫，待比较的字符串指针常规往后，然后比较，如果相等，作为模板的字符串指针往后扫。如果最后比较结果，相等字符数等于模版的长度，就是符合要求的，return true.
AC之后看了discussion，有一个C语言三行AC，非常优雅，在这里放一下。

bool isSubsequence(char* s, char* t) {    while (*t)        s += *s == *t++;    return !*s;}

Complexity analysis

O（n） n为s.size()

Code

*这里一开始搞错了t和s的指代，懒得改了就换了输入的参数名，原型应该是bool isSubsequence(string s, string t)

class Solution {public:    bool isSubsequence(string t, string s) {        int count = 0;        int i = 0, j = 0;        while(i < t.size() && j < s.size()) {                if (s[j] == t[i]) {                    count++;                    i++;                }               j++;            }        if (count == t.size())            return true;        return false;    }};

Result