LeetCode(16) 3Sum Closest

https://sodaoo.github.io/2017/11/21/LeetCode-16/
https://sodaoo.github.io

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Description :

Given an array S of n integers, find three integers in S such that the sum is closest to a given number: target.
Return the sum of the three integers. You may assume that each input would have exactly one solution.

For example, given array S = { -1 2 1 -4 }, target = 1.

The sum that closest to the target is 2. ( cause: -1 + 2 + 1 = 2).

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寻找 array 里面三数之和距离 target 最近的数 .返回这三数之和 .

一开始以为和前面的三数之和有关, 但是很不幸…这是另一个题了 ..

方法就是不断的维护三个数的较小值, 直到遍历结束 .

如何确保不用三重循环? 只需要对 array 排序, 然后用三个指针 ,两个指在最前, 一个指在最后 .

def  threeSumClosest(self,nums,target):   nums.sort()   res = nums[0]+nums[1]+nums[2]  # result   # 两个循环 ,维护相对较小的 res    for i in range(len(nums) - 2):      j , k = i+1 ,len(nums)-1      while j<k:         sum = nums[i]+nums[j]+nums[k]         if sum == target:            return sum         if abs(sum - target) < abs(res - target):            res = sum         if sum < target :            j = j + 1         elif sum > target:            k = k - 1   return res

static int x=[]() {   std::ios::sync_with_stdio(false);   cin.tie(NULL);   return 0;}();class Solution {public:   int threeSumClosest(vector<int>& nums, int target) {      int res = nums[0] + nums[1] + nums[2];      int diff = abs(res - target);      sort(nums.begin(), nums.end());      for (int i = 0; i < nums.size() - 2; ++i) {         int left = i + 1, right = nums.size() - 1;         while (left < right) {            int sum = nums[i] + nums[left] + nums[right];            int newDiff = abs(sum - target);            if (diff > newDiff) {               diff = newDiff;               res = sum;            }            if (sum < target) ++left;            else --right;         }      }      return res;   }};

Thanks for 思路来源