[LeetCode] 3Sum Closest 最近三數之和

Grandyang發表於2015-05-18
LeetCode

 

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}, and target = 1.

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

 

這道題讓我們求最接近給定值的三數之和，是在之前那道 3Sum 三數之和的基礎上又增加了些許難度，那麼這道題讓我們返回這個最接近於給定值的值，即我們要保證當前三數和跟給定值之間的差的絕對值最小，所以我們需要定義一個變數diff用來記錄差的絕對值，然後我們還是要先將陣列排個序，然後開始遍歷陣列，思路跟那道三數之和很相似，都是先確定一個數，然後用兩個指標left和right來滑動尋找另外兩個數，每確定兩個數，我們求出此三數之和，然後算和給定值的差的絕對值存在newDiff中，然後和diff比較並更新diff和結果closest即可，程式碼如下：

 

class Solution {
public:
    int threeSumClosest(vector<int>& nums, int target) {
        int closest = nums[0] + nums[1] + nums[2];
        int diff = abs(closest - 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;
                    closest = sum;
                }
                if (sum < target) ++left;
                else --right;
            }
        }
        return closest;
    }
};

 

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