leetcode76 Minimum Window Substring
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1 """ 2 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). 3 Example: 4 Input: S = "ADOBECODEBANC", T = "ABC" 5 Output: "BANC" 6 """ 7 """ 8 题很难,debug一遍才看明白。debug过程如下: 9 初始化:count{A:1, B:1, C:1} cnt = 0 滑动窗口[] 10 1. {A:0, B:1, C:1} cnt = 1 [A] 11 2. {A:0, B:1, C:1, D:-1} cnt = 1 [AD] 12 3. {A:0, B:1, C:1, D:-1, O:-1} cnt = 1 [ADO] 13 4. {A:0, B:0, C:1, D:-1, O:-1} cnt = 2 [ADOB] 14 5. {A:0, B:0, C:1, D:-1, O:-1, E:-1} cnt = 2 [ADOBE] 15 6. {A:0, B:0, C:0, D:-1, O:-1, E:-1} cnt = 3 [ADOBEC] 16 ----------------cnt==len(t)时将结果保存----------------------- 17 ----------------开始移动窗口左端------------------------------ 18 7. {A:1, B:0, C:0, D:-1, O:-1, E:-1} cnt = 2 [DOBEC] 19 ----------------cnt!=len(t),开始移动窗口右端------------------- 20 8. {A:1, B:0, C:0, D:-1, O:-2, E:-1} cnt = 2 [DOBECO] 21 9. {A:1, B:0, C:0, D:-2, O:-2, E:-1} cnt = 2 [DOBECOD] 22 10. {A:1, B:0, C:0, D:-2, O:-2, E:-2} cnt = 2 [DOBECODE] 23 11. {A:1, B:-1, C:0, D:-2, O:-2, E:-2} cnt = 2 [DOBECODEB] 24 12. {A:0, B:-1, C:0, D:-2, O:-2, E:-2} cnt = 3 [DOBECODEBA] 25 --------------如果小于之前的长度,则保存,移动左端窗口------------ 26 13. {A:0, B:-1, C:0, D:-1, O:-2, E:-2} cnt = 3 [OBECODEBA] 27 14. {A:0, B:-1, C:0, D:-1, O:-1, E:-2} cnt = 3 [BECODEBA] 28 15. {A:0, B:0, C:0, D:-1, O:-1, E:-2} cnt = 3 [ECODEBA] 29 16. {A:0, B:0, C:0, D:-1, O:-1, E:-1} cnt = 3 [CODEBA] 30 17. {A:0, B:0, C:1, D:-1, O:-1, E:-1} cnt = 2 [ODEBA] 31 --------------cnt!=len(t),开始移动窗口右端------------------- 32 18. {A:0, B:0, C:1, D:-1, O:-1, E:-1, N:-1} cnt = 2 [ODEBAN] 33 19. {A:0, B:0, C:0, D:-1, O:-1, E:-1, N:-1} cnt = 3 [ODEBANC] 34 -----------果小于之前的长度,则保存,移动左端窗口------------------ 35 20. {A:0, B:0, C:0, D:-1, O:0, E:-1, N:-1} cnt = 3 [DEBANC] 36 21. {A:0, B:0, C:0, D:0, O:0, E:-1, N:-1} cnt = 3 [EBANC] 37 22. {A:0, B:0, C:0, D:0, O:0, E:0, N:-1} cnt = 3 [BANC] 38 23. {A:0, B:1, C:0, D:0, O:0, E:0, N:-1} cnt = 2 [ANC] 39 """ 40 class Solution: 41 def minWindow(self, s, t): 42 import collections 43 res = "" 44 left, cnt, minLen = 0, 0, float(‘inf‘) 45 count = collections.Counter(t) 46 for i, c in enumerate(s): 47 count[c] -= 1 48 if count[c] >= 0: 49 cnt += 1 50 while cnt == len(t): 51 if minLen > i - left + 1: 52 minLen = i - left + 1 53 res = s[left : i + 1] 54 count[s[left]] += 1 55 if count[s[left]] > 0: 56 cnt -= 1 57 left += 1 58 return res
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