leetcode392. Is Subsequence

题目要求

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?

如何判断字符串s是不是字符串t的一个子序列。子序列是指s中的字母均按照相对位置存在于t中，好比"abc"是"ahbfdc"的一个子序列，可是"axc"就不是"ahbgdc"的一个子序列。

思路一：java API

java中提供了一个String.indexOf(char c, int startIndex)的方法，这个方法是指从字符串中的startIndex位置开始日后找，返回第一个c所在的下标，若是找不到，则返回-1。利用这个方法咱们能够快速的解决这个问题。

public boolean isSubsequence(String s, String t) {
        if(s==null || s.length()==0) return true;
        int start = -1;
        for(int i = 0 ; i<s.length() ; i++) {
            char c = s.charAt(i);
            //从上一个字符的起始位置开始，找下一个字符第一次出现的下标。若是不存在该字符，则说明不是子序列
            start = t.indexOf(c, start+1);
            if(start < 0) return false;
        }
        return true;
    }

思路二：二分法

二分法的思路主要是指，首先咱们遍历字符串t，找到每一个字符在t中出现的位置。当咱们知道每一个字符在t中出现的全部下标后，就开始遍历s，并开始找到距离上一个字符所在的位置以后的当前字符的最小下标。
举例：

s="abc"
t="acbgbc"
遍历t以后能够获得这样一个字段：
a:{0}
b:{2,4}
g:{3}
c:{1,5}

以后遍历s，并用一个index来记录当前字符所在的下标，index初始时为-1。
s[0] = a, a:{0} -> index = 0
s[1] = b, b:{2,4} -> index = 2
s[2] = c, c:{1,5} -> index=5

能够看到咱们可以找到一个合法的序列，使得当前字母的起始下标始终大于上一个字母的下标。

public boolean isSubsequence(String s, String t) {
        List<Integer>[] idx = new List[256]; // Just for clarity
        for (int i = 0; i < t.length(); i++) {
            if (idx[t.charAt(i)] == null)
                idx[t.charAt(i)] = new ArrayList<>();
            idx[t.charAt(i)].add(i);
        }
        
        int prev = 0;
        for (int i = 0; i < s.length(); i++) {
            if (idx[s.charAt(i)] == null) return false; // Note: char of S does NOT exist in T causing NPE
            int j = Collections.binarySearch(idx[s.charAt(i)], prev);
            if (j < 0) j = -j - 1;
            if (j == idx[s.charAt(i)].size()) return false;
            prev = idx[s.charAt(i)].get(j) + 1;
        }
        return true;
    }

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