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). Example: Input: S = "ADOBECODEBANC", T = "ABC" Output: "BANC" Note: If there is no such window in S…
最后更新 二刷 08-Jan-17 回头看了下一刷的,用的map,应该是int[256]的意思,后面没仔细看cuz whatever I was doing at that time.. wasnt good 做法和LC 76非常像,用2 Pointers + 计数来判断是否满足. 这里"有效读取"的判断标准变成了 count[s.charAt(someIndex)]是否从0递增,和每个循环最后它是否递减回0,以此判断dinstinct是否有变化,其实这个比76的有效读取要稍微好理解一…