LeetCode 16. 3sum

文章大纲

• 题目简介与解题思路
• • 题目简介
  • 解法
• c++ 代码
• scala 代码
• python 代码
• 参考文献

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题目简介与解题思路 题目简介

Given an integer array nums of length n and an integer target, find three integers in nums such that the sum is closest to target.

Return the sum of the three integers.

You may assume that each input would have exactly one solution.

Example 1:

Input: nums = [-1,2,1,-4], target = 1
Output: 2
Explanation: The sum that is closest to the target is 2. (-1 + 2 + 1 = 2).
Example 2:

Input: nums = [0,0,0], target = 1
Output: 0

Constraints:

3 <= nums.length <= 1000
-1000 <= nums[i] <= 1000
-104 <= target <= 104

解法

主要的思路是排序后，移动两个指针的思路

Sort the vector and then no need to run O(N^3) algorithm as each index has a direction to move.

The code starts from this formation.

本题的解法不需要

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/-------------------------------------------------------------------
^ ^ -------------------------------------------------------------- ^
| | --------------------------------------------------------------- |
| ± second ------------------------------------------------- third
±first

if nums[first] + nums[second] + nums[third] is smaller than the target, we know we have to increase the sum. so only choice is moving the second index forward.

---

/--------------------------------------------------------------------
^ ^ -------------------------------------------------------------- ^
| | ---------------------------------------------------------------- |
| — ± second ---------------------------------------------- third
±first

if the sum is bigger than the target, we know that we need to reduce the sum. so only choice is moving ‘third’ to backward. of course if the sum equals to target, we can immediately return the sum.

---

/----------------------------------------------------------------------------
^ ^ -------------------------------------------------------------- ^
| | ---------------------------------------------------------------- |
| ± second ------------------------------------------------------- third
±first

when second and third cross, the round is done so start next round by moving ‘first’ and resetting second and third.

---

/---------------------------------------------------------------------
— ^ — ^--------------------------------------------------------- ^
— | ---- | --------------------------------------------------------- |
— | ---- ± second ------------------------------------------- third
— ±first

while doing this, collect the closest sum of each stage by calculating and comparing delta. Compare abs(target-newSum) and abs(target-closest). At the end of the process the three indexes will eventually be gathered at the end of the array.

---

----------------------------------------------------------------- ^ ^ ^
------------------------------------------------------------------ | | `- third
------------------------------------------------------------------ | ± second
------------------------------------------------------------------ ±first

if no exactly matching sum has been found so far, the value in closest will be the answer.

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c++ 代码

class Solution {
public:
    int threeSumClosest(vector<int>& nums, int target) {
        sort(nums.begin(), nums.end());
        int n = nums.size(), ans = nums[0] + nums[1] + nums[2];
        for (int i = 0; i < n-2; ++i) {
            int l = i + 1, r = n - 1;
            while (l < r) {
                int sum3 = nums[i] + nums[l] + nums[r];
                if (abs(ans - target) > abs(sum3 - target)) // Update better ans
                    ans = sum3;
                if (sum3 == target) break;
                if (sum3 > target)
                    --r; // Sum3 is greater than the target, to minimize the difference, we have to decrease our sum3
                else
                    ++l; // Sum3 is lesser than the target, to minimize the difference, we have to increase our sum3
            }
        }
        return ans;
    }
};

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scala 代码

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python 代码

class Solution:
    def threeSumClosest(self, nums: [int], target: int) -> int:

        nums.sort()
        result = sum(nums[:3])

        for i in range(len(nums)):
            left,right = i+1,len(nums)-1
            while left < right:
                s = sum((nums[i],nums[left],nums[right]))
                if abs(s - target) < abs(result - target):
                    result = s
                if s < target:
                    left = left +1
                elif s> target:
                    right = right -1
                else:
                    return result

        return  result



if __name__ == "__main__":
    s = Solution()
    print(s.threeSumClosest([-1,2,1,-4],1))

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参考文献

https://leetcode.com/problems/3sum-closest/

scala 版本题解
https://leetcode.com/problems/3sum-closest/discuss/?currentPage=1&orderBy=most_votes&query=&tag=scala

带有解释的题解
https://leetcode.com/problems/3sum-closest/discuss/7883/C%2B%2B-solution-O(n2)-using-sort

python 题解 2 个指针

https://leetcode.com/problems/3sum-closest/discuss/8026/Python-solution-(two-pointer).