【030-Substring with Concatenation of All Words（串联全部单词的子串）】

【LeetCode-面试算法经典-Java实现】【全部题目文件夹索引】

原题

　　You are given a string, s, and a list of words, words, that are all of the same length. Find all starting indices of substring(s) in s that is a concatenation of each word in words exactly once and without any intervening characters.

　　For example, given:

　　s: "barfoothefoobarman"

　　words: ["foo", "bar"]

　　You should return the indices: [0,9].

　　(order does not matter).

题目大意

　　给定一个字符串s和一个字符串数组words，wrods中的字符串长度都相等。找出s中全部的子串恰好包括words中全部字符各一次，返回子串的起始位置。

解题思路

　　把words转化为一个HashMap

代码实现

算法实现类

import java.util.*;

public class Solution {

    public List<Integer> findSubstring(String s, String[] words) {
        List<Integer> list = new ArrayList<Integer>();
        if (words.length == 0) return list;
        int wLen = words[0].length();
        int len = s.length();
        if (len < wLen * words.length) return list;
        Map<String, Integer> mapW = new HashMap<String, Integer>();
        for (String word : words)
            mapW.put(word, mapW.containsKey(word) ? mapW.get(word) + 1 : 1);
        for (int start = 0; start < wLen; start++) {
            int pos = start;
            int tStart = -1;
            Map<String, Integer> mapT = new HashMap<String, Integer>(mapW);
            while (pos + wLen <= len) {
                String cand = s.substring(pos, pos + wLen);
                if (!mapW.containsKey(cand)) {
                    if (tStart != -1) mapT = new HashMap<String, Integer>(mapW);
                    tStart = -1;
                } else if (mapT.containsKey(cand)) {
                    tStart = tStart == -1 ?

pos : tStart;
                    if (mapT.get(cand) == 1) mapT.remove(cand);
                    else mapT.put(cand, mapT.get(cand) - 1);
                    if (mapT.isEmpty()) list.add(tStart);
                } else {
                    while (tStart < pos) {
                        String rCand = s.substring(tStart, tStart + wLen);
                        if (cand.equals(rCand)) {
                            tStart += wLen;
                            if (mapT.isEmpty()) list.add(tStart);
                            break;
                        }
                        tStart += wLen;
                        mapT.put(rCand, mapT.containsKey(rCand) ? mapT.get(rCand) + 1 : 1);
                    }
                }
                pos += wLen;
            }
        }
        return list;
    }
}

评測结果

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特别说明

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