Problem

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

题意

给出两个字符串s和t，要求判断s是否是t的子串。

该题中对于子串的定义是：只要s中的字母以相同的顺序出现在t中，那么我们就说s是t的子串

示例见原题。

分析

从贪心算法的角度去看，解决这道题的思路是每判断一次都得到一个目前看起来最优的解，那么对于这道题来说，就要求每判断一次都能将问题的规模缩小。

所以解法就是。从s的第一个字母s[0]开始，从t[0]开始逐个匹配t中的字母，如果在t[i]处匹配，则继续匹配s[1]，从t[i+1]开始。

重复这个过程，直到s中的字母全部完成匹配，则返回true；否则，到达t[t.size()-1]时仍未完成s的匹配，返回false。

Code

//Runtime: 68ms
class Solution {
public:
    bool isSubsequence(string s, string t) {
        int sIndex = 0, tIndex = 0;             //index of string s and string t
        int sSize = s.size(), tSize = t.size();
        if (sSize == 0) return true;         //if s is empty string, s is every string's sub-string
        while (tIndex < tSize) {                //check all characters of string t
            if (s[sIndex] == t[tIndex]) {
                sIndex++;                       //if s[sIndex] and t[tIndex] are matched, then check the next character of s
                if (sIndex == sSize)            //which means all characters of s are matched
                    return true;
            }
            tIndex++;
        }
        return false;
    }
};