leetcode-5.22[*748. 最短完整詞、653. 兩數之和 IV - 輸入 BST、821. 字符的最短距離]（python實現)

題目1

題解1

class Solution:
    def shortestCompletingWord(self, licensePlate: str, words: List[str]) -> str:
        s = [w for w in licensePlate.lower() if ord(w)>=97 and ord(w)<=122]
        words.sort(key=lambda x:len(x))   #按單詞長度進行排序
        for w in words: 
            w_copy=list(w)
            for i in s:             #判斷單詞是否包含牌照所有字母
                if i in w_copy:
                    w_copy.remove(i)
                else:                    #不滿足條件
                    break
            else:
                return w                    #返回滿足條件的單詞

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題目2

題解2

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def findTarget(self, root: TreeNode, k: int) -> bool:
        num = self.getSortlist(root)
        l,r = 0, len(num)-1
        while l < r:
            if num[l] + num[r] < k:
                l += 1
            elif num[l] + num[r] > k:
                r -= 1
            else:
                return True
        return False


    def getSortlist(self, root):
        if root is None:
            return []
        return self.getSortlist(root.left) + [root.val] + self.getSortlist(root.right)

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題目3

題解3

class Solution:
    def shortestToChar(self, S: str, C: str) -> List[int]:
        """
            正向求每個字符距離C的距離
        """
        # 默認的prev無窮大
        prev = float('-inf')
        ans = []
        for i, x in enumerate(S):
            if x == C: prev = i
            ans.append(i - prev)

        """
            反向求每個字符距離C的距離
        """
        prev = float('inf') 
        for i in range(len(S) - 1, -1, -1):
            if S[i] == C: prev = i
            # 求出對應座標距離的最小值
            ans[i] = min(ans[i], prev - i)
        return ans

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