Leetcode 第18题: 4Sum(4个数相加)

题目地址:4Sum


题目简介:

给定一个包含n个整数的数组nums和一个目标值target,判断nums中是否存在四个元素a,b,c,d,使得a+b+c+d的值与target相等?找出所有满足条件且不重复的四元组。

注意答案中不可以包含重复的四元组。

Given array nums = [1, 0, -1, 0, -2, 2], and target = 0.

A solution set is:
[
  [-1,  0, 0, 1],
  [-2, -1, 1, 2],
  [-2,  0, 0, 2]
]

题目解析:3数相加的加强版,那就老法子走新路

C++版:

class Solution {
public:
    vector<vector<int>> fourSum(vector<int>& nums, int target) {
        vector<vector<int>> ans_set;
        if (nums.size() <= 3)
        {
            return ans_set;   
        }
        sort(nums.begin(), nums.end());
        
        for(int j = 0; j <= nums.size() - 4; j++)
        {
            if (j > 0 && nums[j] == nums[j - 1])    continue;
            for (int i = j + 1; i < nums.size() - 2; i++)
            {
                if (i > j + 1 && nums[i] == nums[i - 1])    continue;
                int l = i + 1, r = nums.size() - 1;
                while(l < r)
                {
                    if (nums[i] + nums[l] + nums[r] > target - nums[j])    r--;
                    else if (nums[i] + nums[l] + nums[r] < target - nums[j])   l++;
                    else
                    {
                        ans_set.push_back({nums[j],nums[i], nums[l], nums[r]});
                        while(l < r && nums[l] == nums[l + 1])  l++;
                        while(l < r && nums[r] == nums[r - 1])  r--;
                        l++;
                        r--;
                    }  
                }
            }
        }
        return ans_set;
    }
};

Python版:

class Solution:
    def fourSum(self, nums: List[int], target: int) -> List[List[int]]:
        ans_list = []
        if (len(nums) <= 3): #如果当前的数字少于三个那么直接返回结果
            return (ans_list)

        nums.sort()

        for j in range(0,len(nums) - 3):
            if (j > 0 and nums[j] == nums[j - 1]):
                    continue
            for i in range(j + 1,len(nums) - 2):
                if (i > j + 1 and nums[i] == nums[i - 1]):
                    continue
                l = i + 1
                r = len(nums) - 1
                while (l < r):
                    if (nums[i] + nums[l] + nums[r] + nums[j] == target):
                        ans_list.append([nums[j],nums[i],nums[l],nums[r]])
                        while (l < r and nums[l] == nums[l + 1]):
                            l += 1
                        while (l < r and nums[r] == nums[r - 1]):
                            r -= 1
                        l += 1
                        r -= 1
                    elif (nums[i] + nums[l] + nums[r] > target - nums[j]):
                        r -=1
                    else:
                        l += 1
        return (ans_list)

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转载自blog.csdn.net/chao_shine/article/details/88564021