lintcode: 三数之和II

题目

三数之和 II

给一个包含n个整数的数组S, 找到和与给定整数target最接近的三元组，返回这三个数的和。

样例

例如S = [-1, 2, 1, -4] and target = 1.  和最接近1的三元组是 -1 + 2 + 1 = 2.

注意

只需要返回三元组之和，无需返回三元组本身

解题

和上一题差不多，程序也只是稍微修改了

public class Solution {
    /**
     * @param numbers: Give an array numbers of n integer
     * @param target : An integer
     * @return : return the sum of the three integers, the sum closest target.
     */
    public int threeSumClosest(int[] numbers ,int target) {
        // write your code here
        if(numbers == null)
            return 0;
        Arrays.sort(numbers);
        int sum = Integer.MAX_VALUE ;
        int numlen = numbers.length;
        for( int i = 0;i< numlen ;i++){
            int left = i + 1;
            int right = numlen - 1;
            while(left < right){
                int tmpsum = numbers[i] + numbers[left] + numbers[right];
                int tmpdist = tmpsum - target;
                sum = Math.abs(tmpdist) > Math.abs(sum - target) ? sum:tmpsum;
                if(tmpdist == 0){
                    return target;
                }
                if(tmpdist <0){
                    left++;
                }else{
                    right--;
                }

            }
        }
        return sum;
    }
}

总耗时: 1504 ms

class Solution:
    """
    @param numbers: Give an array numbers of n integer
    @param target : An integer
    @return : return the sum of the three integers, the sum closest target.
    """
    def threeSumClosest(self, numbers, target):
        # write your code here
        if numbers == None:
            return 0 
        numlen = len(numbers)
        numbers.sort()
        sum = 0
        for i in range(numlen-1):
            left = i + 1
            right = numlen - 1
            while left < right:
                tmpsum = numbers[i] + numbers[left] + numbers[right]
                tmpdist = tmpsum - target
                if i==0:
                    sum = tmpsum
                sum = sum if abs(tmpdist) > abs(sum-target) else tmpsum 
                if tmpdist == 0:
                    return target
                if tmpdist < 0:
                    left +=1
                else:
                    right -=1
        return sum

总耗时: 403 ms