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添加1,395字节 、 2021年6月15日 (二) 19:29
无编辑摘要
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== 使用 pattern ==
 
== 使用 pattern ==
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<ref>[http://www.noteanddata.com/leetcode-3-Longest-Substring-Without-Repeating-Characters-java-sliding-windows-solution-note.html#sliding-windows滑动窗口的特征和基本模版 sliding windows 滑动窗口的特征和基本模版]</ref> <ref>https://blog.csdn.net/fuxuemingzhu/article/details/82931106</ref>
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在需要找一个区间的问题中,首先移动右指针使得满足要求,再移动左指针直到不满足要求,再移动右指针,持续如此,直到区间的右边界到达整体的结束点。
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在需要找一个区间的问题中,
<ref>[http://www.noteanddata.com/leetcode-3-Longest-Substring-Without-Repeating-Characters-java-sliding-windows-solution-note.html#sliding-windows滑动窗口的特征和基本模版 sliding windows 滑动窗口的特征和基本模版]</ref> <ref>https://blog.csdn.net/fuxuemingzhu/article/details/82931106</ref>
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'''第一种情况:'''首先移动右指针使得满足要求,再移动左指针直到不满足要求,再移动右指针,持续如此,直到区间的右边界到达整体的结束点。
 
<syntaxhighlight lang=python>
 
<syntaxhighlight lang=python>
 
left = 0
 
left = 0
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init state
 
ret = worse case retval
 
ret = worse case retval
 
for right in range(n):
 
for right in range(n):
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         left += 1
 
         left += 1
 
         update state
 
         update state
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return ret
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</syntaxhighlight>
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'''第二种情况:'''首先移动右指针直到不满足要求,再移动左指针直到满足要求,此时记录当前状态,若更优则记录下来。持续如此,直到区间的右边界到达整体的结束点。
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<syntaxhighlight lang=python>
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left = 0
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init state
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ret = worse case retval
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for right in range(n):
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    update state
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    while state is not valid:
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        left += 1
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        update state
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    sol = f(left, right)
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    if sol is better than ret:
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        ret = sol
 
return ret
 
return ret
 
</syntaxhighlight>
 
</syntaxhighlight>
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=== [https://leetcode.com/problems/minimum-window-substring/ LeetCode 76. Minimum Window Substring] ===
 
=== [https://leetcode.com/problems/minimum-window-substring/ LeetCode 76. Minimum Window Substring] ===
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Python 解法:
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第一种情况。Python 解法:
    
<syntaxhighlight lang=python>
 
<syntaxhighlight lang=python>
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             if chars[sl] < target[sl]:
 
             if chars[sl] < target[sl]:
 
                 cnt -= 1
 
                 cnt -= 1
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    return ret
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</syntaxhighlight>
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 +
=== LeetCode 340. Longest Substring with At Most K Distinct Characters ===
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 +
第二种情况。Python 解法:
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 +
<syntaxhighlight lang=python>
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def lengthOfLongestSubstringKDistinct(self, s: str, k: int) -> int:
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    left = 0
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    ret = 0
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    distinct = 0
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    chars = Counter()
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    for right, ch in enumerate(s):
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        # move the right pointer until the state is invalid
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        if chars[ch] == 0:
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            distinct += 1
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        chars[ch] += 1
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        while distinct > k and left <= right:
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            # move the left pointer until the state is valid
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            c = s[left]
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            left += 1
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            chars[c] -= 1
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            if chars[c] == 0:
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                distinct -= 1
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        # after the state becomes valid, record the best result
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        ret = max(right - left + 1, ret)
 
     return ret
 
     return ret
 
</syntaxhighlight>
 
</syntaxhighlight>