leetcode第16題——**3Sum Closest

題目

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).

思路

這道題跟3Sum那道題（請看本人的上一篇博客）思路類似，先固定一個數，然後用頭尾指針從兩端向中間靠攏，每次遇到sum更接近target的時候就更新結果，算法的時間複雜度爲O(n^2)，運行結果擊敗了9474%的人 : D。

代碼

Python

class Solution(object):
    def threeSumClosest(self, nums, target):
        """
        :type nums: List[int]
        :type target: int
        :rtype: int
        """
        leng = len(nums)
        if (leng < 3):
            return 0
        nums.sort()
        res = nums[0] + nums[1] + nums[2]
        minSub = abs(res - target)
        for i in xrange(leng):
            if (i > 0 and nums[i-1] == nums[i]):
                continue
            p = i + 1
            q = leng - 1
            while (p < q):
                tsum = nums[i] + nums[p] + nums[q]
                sub = tsum - target
                if (sub == 0):
                    return target
                if (abs(sub) < minSub):
                    #如果和更接近target就更新結果
                    res = tsum
                    minSub = abs(sub)
                    
                if (sub < 0):
                    p += 1
                else:
                    q -= 1
        return res

Java

public class Solution {
    public int threeSumClosest(int[] nums, int target) {
        int len = nums.length;
        if (nums == null || len < 3) return 0;
        //先對數組排序
        Arrays.sort(nums);
        
        
        int i,p,q,sum,sub;
        int res = nums[0] + nums[1] + nums[2];
        int minSub = Math.abs(res - target);
        
        for (i = 0;i < len;i++){
        	if (i > 0 && nums[i] == nums[i-1]) continue;
        	
        	p = i + 1;
        	q = len - 1;
        	while (p < q){
        		sum = nums[i] + nums[p] + nums[q];
        		sub = sum - target;
        		
        		if (sub == 0) return target;
        		//如果有更接近target的，更新結果
        		if (Math.abs(sub) < minSub){
        			res = sum;
        			minSub = Math.abs(sub);
        		}
        		
        		if (sub < 0) p++;
        		else q--;
        	}
        }
        
        return res;
    }
}