LeetCode刷題記錄11-15

11. 盛最多水的容器

題解

    left_h = 0
    right_h = -1
    sc = len(height)-1
    max_s= 0
    while sc != 0:
        if height[left_h] > height[right_h]:
            area = sc * height[right_h]
            right_h -= 1
        else:
            area = sc * height[left_h]
            left_h += 1
        sc -= 1
        max_s=max(area,max_s)
    return max_s

15. 三數之和

題解

       n = len(nums)
        if not nums or n < 3:
            return []
        nums.sort()
        res = []
        for i in range(n):
            if nums[i] > 0:
                return res
            if i > 0 and nums[i] == nums[i - 1]:
                continue
            L = i + 1  # 令左指針 L=i+1L=i+1
            R = n - 1  # 令右指針 R=n-1R=n−1
            while L < R: # 當 L<R 時，執行循環
                if nums[i] + nums[L] + nums[R] == 0:  # 判斷左界和右界是否和下一位置重複，去除重複解。並同時將 L,R 移到下一位置，尋找新的解
                    res.append([nums[i], nums[L], nums[R]])
                    while L < R and nums[L] == nums[L + 1]:
                        L = L + 1
                    while L < R and nums[R] == nums[R - 1]:
                        R = R - 1
                    L = L + 1
                    R = R - 1
                elif nums[i] + nums[L] + nums[R] > 0: # 若和大於0,說明nums[R]太大,R左移
                    R = R - 1
                else:         #若和小於 0，說明 nums[L]太小，L 右移
                    L = L + 1
        return res

思路非常棒的解答

19. 刪除鏈表的倒數第N個節點

題解

        root = ListNode(-1)
        root.next = head
        left = right = root
        while n:
            right = right.next
            n -= 1

        while right.next:
            right = right.next
            left = left.next

        left.next = left.next.next

        return root.next

20. 有效的括號

題解

        dic = {'{': '}',  '[': ']', '(': ')', '?': '?'}
        stack = ['?']
        for c in s:
            if c in dic: stack.append(c)
            elif dic[stack.pop()] != c: return False
        return len(stack) == 1

21. 合併兩個有序鏈表

題解

        if l1 is None:
            return l2

        if l2 is None:
            return l1

        if l1.val < l2.val:
            l1.next = self.mergeTwoLists(l1.next, l2)
            return l1
        else:
            l2.next = self.mergeTwoLists(l1, l2.next)
            return l2