LeetCode 392. Is Subsequence (判断子序列)

题目标签：Greedy

　　设两个 pointers，s_index  和  t_index；

　　如果 s_index  和  t_index 位置上的字母一样，那么继续移动两个 pointers；

　　如果字母不一样，只移动 t_index；

　　具体看code。

Java Solution:

Runtime:  7 ms, faster than 80.12% 

Memory Usage: 44.5 MB, less than 100.00%

完成日期：02/19/2020

关键点：two pointers

class Solution {
    public boolean isSubsequence(String s, String t) {
        if(s.length() == 0) 
            return true;
        
        int s_index =0, t_index = 0;
        
        // use two pointers to find char in s and t
        while(t_index < t.length()) {
            if(s.charAt(s_index) == t.charAt(t_index)) { // if found the match, move s_index to next one
                s_index++;
                if(s_index == s.length())
                    return true;
            }
            t_index++;
        }
        return false;
    }
}

参考资料：LeetCode Discuss

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