LeetCode算法题python解法：15. 3Sum 16. 3Sum Closest 18. 4Sum

三数之和与四数之和算法解析

最新推荐文章于 2025-07-20 10:41:25 发布

转载最新推荐文章于 2025-07-20 10:41:25 发布 · 102 阅读

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原文链接：http://www.cnblogs.com/slarker/p/9657924.html

本文详细解析了求解三数之和和四数之和的算法，通过夹逼定理实现对目标值的逼近，提供Python代码实现。适用于面试准备和算法学习。

这三个题目基本属于同一类型，这里我的解题思路是所谓的‘夹逼定理’。不同的题目运用该思路会有一点区别，总体上是大同小异。

先来看15题：

题目要求取出数列中可以加起来为0的三个数来组成序列并添加到新的序列中去。这个题目其实非常简单，和之前水桶装水的题目有些类似。这里只需要先绑定两端，中间的值动。

class Solution(object):
    def threeSum(self, nums):
        nums = sorted(nums)
        print(nums)
        out = []
        for i in range(len(nums) - 2):
            if i>0 and nums[i] == nums[i-1]:
                continue
            f = i + 1
            b = len(nums) - 1
            while True:
                if f >= b:
                    break
                if nums[i] + nums[f] + nums[b] < 0:
                    f += 1
                elif nums[i] + nums[f] + nums[b] > 0:
                    b -= 1
                else:
                    out.append([nums[i], nums[f], nums[b]])
                    # if sorted([nums[i], nums[f], nums[b]]) not in out:
                    #    out.append(sorted([nums[i], nums[f], nums[b]]))
                    f += 1
                    b -= 1
                    while f + 1 <= b and nums[f] == nums[f - 1]:
                        f += 1
                    while b - 1 >= f and nums[b] == nums[b + 1]:
                        b -= 1

        return out

16.

class Solution:
    def threeSumClosest(self, nums, target):
        nums = sorted(nums)
        disminx = abs(nums[1] - nums[-1])+ abs(target)
        minx = sum(nums[:3])

        for i in range(len(nums)):
            f, b = i + 1, len(nums) - 1

            while f < b:
                num = nums[i] + nums[f] + nums[b]
                if num == target:
                    return target
                elif num > target:
                    b -= 1
                    minn = abs(num - target)
                    if disminx > minn:
                        disminx = minn
                        minx = num

                else:
                    f += 1
                    minn = abs(num - target)
                    if disminx > minn:
                        disminx = minn
                        minx = num
        return minx

class Solution:
    def fourSum(self, nums, target):
        out = []
        nums = sorted(nums)

        for i in range(len(nums)-3):
            if i > 0 and nums[i] == nums[i - 1]:
                continue

            for b in range(len(nums)-1,i+2,-1):
                if b < len(nums)-1 and nums[b] == nums[b+1]:
                    continue
                f = i + 1
                bf = b - 1
                while True:
                    Sum = nums[i] + nums[f] + nums[bf] + nums[b]
                    #print(i, f, bf, b)
                    #print([nums[i], nums[f], nums[bf], nums[b]])
                    if bf <= f:
                        break
                    elif Sum == target:
                        print('==')
                        out.append([nums[i],nums[f],nums[bf] , nums[b]])
                        bf-=1
                        while  0< bf < len(nums)-1 and nums[bf+1] == nums[bf]:
                            bf -= 1
                        f+=1
                        while len(nums) -1> f >0 and nums[f] == nums[f-1]:
                            f+=1
                    elif Sum > target:
                        bf-=1
                    elif Sum < target:
                        f += 1
        return out