【030-Substring with Concatenation of All Words（串联所有单词的子串）】【LeetCode-面试算法经典-Java实现】【所有题目目录索引】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中所有的子串恰好包含words中所有字符各一次，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;}}评测结果
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