LeetCode:Minimum Window Substring

发布时间:2016-12-10 8:03:27 编辑:www.fx114.net 分享查询网我要评论
本篇文章主要介绍了"LeetCode:Minimum Window Substring",主要涉及到LeetCode:Minimum Window Substring方面的内容,对于LeetCode:Minimum Window Substring感兴趣的同学可以参考一下。

Given a string S and a string T, find the minimum window in S which will contain all the characters in T in complexity O(n). For example, S = "ADOBECODEBANC" T = "ABC" Minimum window is "BANC". Note: If there is no such window in S that covers all characters in T, return the emtpy string "". If there are multiple such windows, you are guaranteed that there will always be only one unique minimum window in S. 这题很有意思,是LeetCode上最有意思的题目之一。 我的解法时间复杂度是O(N)。我没想明白需要怎么描述我的解法,就不写了。 需要注意S="a",T="aa"的情况。 package leetcode; import java.util.Arrays; import java.util.HashMap; import java.util.LinkedList; public class MinimumWindowSubstring { public static void main(String[] args) { System.out.println(new MinimumWindowSubstring().minWindow("ADOBECODEBANC", "ABC")); } public String minWindow(String S, String T) { if (T.length() > S.length()) { return ""; } //各字符在T中出现的次数 HashMap<Character, Integer> countMap = new HashMap<Character, Integer>(); for (int i = 0; i < T.length(); i++) { Integer count = countMap.get(T.charAt(i)); if (count == null) { count = 0; } countMap.put(T.charAt(i), count+1); } //搜索S时,保存出现的字符和索引,KEY是字符,List保存字符在S中的索引, //因为T中可能出现重复字符的情况,如S="aa",T="aa"。同一字符可能需要多次匹配,故将其在S中出现的坐标需要保存在List中 HashMap<Character, LinkedList<Integer>> map = new HashMap<Character, LinkedList<Integer>>(); //引入indexMap和indexQueue,以方便查找minIndex boolean[] indexMap = new boolean[S.length()];//标记S中的字符是否出现当前待选匹配中 Arrays.fill(indexMap, false); LinkedList<Integer> indexQueue = new LinkedList<Integer>();//保存S中出现的所有T中字符 int index = 0; int minLen = Integer.MAX_VALUE; int minIndex = -1; int minStart = -1; int windowCount = 0; while (index < S.length()) { char c = S.charAt(index); Integer count = countMap.get(c); if (count != null) { LinkedList<Integer> indexList = map.get(c); if (indexList == null) { indexList = new LinkedList<Integer>(); map.put(c, indexList); } if (indexList.size() == count) { indexList.add(index); Integer firstIndex = indexList.getFirst(); indexList.removeFirst(); indexMap[firstIndex] = false; indexMap[index] = true; indexQueue.addLast(index); //update minIndex if (firstIndex == minIndex) { Integer inx = indexQueue.getFirst(); while (!indexMap[inx]) { indexQueue.removeFirst(); inx = indexQueue.getFirst(); } minIndex = inx; } } else { indexList.add(index); if (minIndex == -1) { minIndex = index; } windowCount++; indexMap[index] = true; indexQueue.addLast(index); } if (windowCount == T.length()) { //get an answer int len = index - minIndex + 1; if (len < minLen) { minLen = len; minStart = minIndex; } } } index++; } if (minStart == -1) { return ""; } return S.substring(minStart, minStart + minLen); } }

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