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. 解题思路: 这是比较难的一道题目,如果t中没...
problem:https://leetcode.com/problems/minimum-window-substring/ 滑动窗口题。维护一个包含t中所有字符的最小滑动窗口,首先用一个hashmap记录所有t中的字符和出现次数,在s中每遇到一次计数器加一,找到了符合条件的窗口后,尝试向右移动窗口左指针,直到恰好能够满足条件为止。更新当前最小滑动窗口。 classSolution {pu...
http://www.lintcode.com/en/problem/minimum-window-substring/ 【题目解析】 可以用窗口型两个指针的思路来解决,外层for循环i = 0 ... n, 内层while循环,条件是j < source.length() 和 !isValid(sourceHash, targetHash),前者是数组下标边界,后者是一个函数,检查当前窗口中的substring是否包含了目标字符串...
Can you solve this real interview question? Minimum Window Substring - Given two strings s and t of lengths m and n respectively, return the minimum window substring of s such that every character in t (including duplicates) is included in the window. If
Minimum Window Substring 这题还是很有难度的。通过数组记录元素的长度,将目标数组与其进行比较,再对相应的子串进行缩小,并记录相应的起点与长度。 提取包含子串的最小窗口,可以是无序的。 1、建立一个256个大小的ASCII数组,统计子串中每个字符出现的次数; 2、用两个指针,一个指针表示窗口的起始位置,一个不断后...
http://www.lintcode.com/en/problem/minimum-window-substring/ 【题目解析】 可以用窗口型两个指针的思路来解决,外层for循环i = 0 ... n, 内层while循环,条件是j < source.length() 和 !isValid(sourceHash, targetHash),前者是数组下标边界,后者是一个函数,检查当前窗口中的substring是否包含了目标字符串...
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