【HOT100】DAY2 - 技术栈

盛最多水的容器

双指针，通过指针大小控制循环；更小的高参与迭代。

python 复制代码

class Solution:
    def maxArea(self, height: List[int]) -> int:
        left = 0
        right = len(height) - 1
        max_area = 0
        while left < right:
            high = min(height[left], height[right])
            width = right - left
            current_area = high * width
            max_area = max(current_area, max_area)
            if height[left] < height[right]:
                left += 1
            else:
                right -= 1
        return max_area

三数之和

循环，i作为一个指针；循环内双指针，左右分别剪枝/去重；通过三数之和的大小决定双指针移动方向

python 复制代码

class Solution:
    def threeSum(self, nums: list[int]) -> list[list[int]]:
        n = len(nums)
        res = []
        if n < 3:
            return res
        
        nums.sort()
        for i in range(n):
            if nums[i] > 0:
                break
            if i > 0 and nums[i] == nums[i - 1]:
                continue
            
            left = i + 1
            right = n - 1
            while left < right:
                sum_three = nums[i] + nums[left] + nums[right]
                if sum_three == 0:
                    res.append([nums[i], nums[left], nums[right]])
                    while left < right and nums[left] == nums[left + 1]:
                        left += 1
                    while left < right and nums[right] == nums[right - 1]:
                        right -= 1
                    left += 1
                    right -= 1
                elif sum_three < 0:
                    left += 1
                else:
                    right -= 1
        return res

接雨水

每次只做一步动作，接水和移动指针错开，左右也错开。移动的下一步要是比当前大，就更新大小，若比当前小，二者差值就是雨水数量。

python 复制代码

class Solution:
    def trap(self, height: List[int]) -> int:
        n = len(height)
        if n < 3:
            return 0
        
        left = 0
        right = n - 1
        left_max = 0
        right_max = 0
        total_water = 0

        while left < right:
            if height[left] < height[right]:
                if height[left] >= left_max:
                    left_max = height[left]
                else:
                    total_water += left_max - height[left]
                left += 1
            else:
                if height[right] >= right_max:
                    right_max = height[right]
                else:
                    total_water += right_max - height[right]
                right -= 1
        
        return total_water

无重复字符的最长子串

哈希+滑动窗口，用s的序号当作右指针并参与循环。

python 复制代码

class Solution:
    def lengthOfLongestSubstring(self, s: str) -> int:
        hash_map = {}
        max_length = 0
        left = 0
        for right, char in enumerate(s):
            if char in hash_map and hash_map[char] >= left:
                left = hash_map[char] + 1
            hash_map[char] = right
            current_length = right - left + 1
            max_length = max(current_length, max_length)
        return max_length