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  • 【python刷题】滑动窗口法

    模板

    left,right = 0,0
    while right < len(s):
        windows.append(s[right])
        right += 1
        while (windows needs shrink):
            window.pop(0)
            left += 1
    

    leetcode 76 最小覆盖子串

    class Solution:
        def minWindow(self, s: str, t: str) -> str:
            from collections import Counter
            needs = Counter(t)
            windows = {i:0 for i in t}
            left, right = 0, 0
            valid = 0
            # 记录最小覆盖子串的起始位置以及长度
            start = 0
            leng = float("inf")
            while right < len(s):
                c = s[right]
                right += 1
                if c in needs:
                    windows[c] += 1
                    if needs[c] == windows[c]:
                        valid += 1
                while valid == len(needs):
                    if right - left < leng:
                        start = left
                        leng = right - left
                    d = s[left]
                    left += 1
                    if d in needs:
                        if windows[d] == needs[d]:
                            valid -= 1
                        windows[d] -= 1
            return "" if leng == float("inf") else s[start:start+leng]
    

    leetcode 567 字符串排列

    class Solution:
        def checkInclusion(self, s1: str, s2: str) -> bool:
            from collections import Counter
            needs = Counter(s1)
            windows = {i:0 for i in s1}
            left, right = 0, 0
            valid = 0
            while right < len(s2):
                c = s2[right]
                right += 1
                if c in needs:
                    windows[c] += 1
                    if needs[c] == windows[c]:
                        valid += 1
                while right - left > len(s1) - 1: # 窗口中只能有len(s1)个元素,
                    if valid == len(needs):
                        print(s2[left:right])
                        return True
                    d = s2[left]
                    left += 1
                    if d in needs:
                        if windows[d] == needs[d]:
                            valid -= 1
                        windows[d] -= 1
            return False
    

    leetcode 438 找所有字母异位词

    class Solution:
        def findAnagrams(self, s: str, p: str) -> List[int]:
            from collections import Counter
            needs = Counter(p)
            windows = {i:0 for i in p}
            left, right = 0, 0
            valid = 0
            res = []
            while right < len(s):
                c = s[right]
                right += 1
                if c in needs:
                    windows[c] += 1
                    if needs[c] == windows[c]:
                        valid += 1
                while right - left > len(p) - 1: # 窗口中只能有两个元素,
                    if valid == len(needs):
                        print(s[left:right])
                        res.append(left)
                    d = s[left]
                    left += 1
                    if d in needs:
                        if windows[d] == needs[d]:
                            valid -= 1
                        windows[d] -= 1
            return res
    

    leetcode 3 最长无重复子串

    class Solution:
        def lengthOfLongestSubstring(self, s: str) -> int:
            if len(s) == 0:
                return 0
            if len(s) == 1:
                return 1
            from collections import defaultdict
            windows = defaultdict(int)
            left, right = 0,0
            res = float("-inf")
            while right < len(s):
                c = s[right]
                right += 1
                windows[c] += 1
                while windows[c] > 1:
                    d = s[left]
                    left += 1
                    windows[d] -= 1
                res = max(res, right - left)
            return res
    
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  • 原文地址:https://www.cnblogs.com/xiximayou/p/14368112.html
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