【一次过】Lintcode 59. 最接近的三数之和

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1:

输入:[2,7,11,15],3
输出:20
解释:
2+7+11=20

例2:

输入:[-1,2,1,-4],1
输出:2
解释:
-1+2+1=2

挑战

O(n^2) 时间, O(1) 额外空间。

注意事项

只需要返回三元组之和，无需返回三元组本身

---

解题思路1：

标准DFS解法。需要增加sum变量来表示截至递归到当前层的和

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
        res = Integer.MAX_VALUE;
        
        dfs(numbers, target, new ArrayList<Integer>(), 0, 0);
        
        return res;
    }
    
    private int res;
    
    private void dfs(int[] numbers, int target, List<Integer> list, int sum, int index){
        if(list.size() == 3){
            if(Math.abs(sum-target) < Math.abs(res-target))
                res = sum;
            
            return;
        }
        
        for(int i=index; i<numbers.length; i++){
            list.add(numbers[i]);
            
            dfs(numbers, target, list, sum+numbers[i], i+1);
            
            list.remove(list.size()-1);
        }
    }
}

解题思路2：

同Lintcode 57思路2。只是需要增加diff变量统计最小差。

class Solution {
    public int threeSumClosest(int[] nums, int target) {
        Arrays.sort(nums);
        int res = 0;
        int diff = Integer.MAX_VALUE;

        for(int i=0; i<nums.length-2; i++){
            int curTarget = target - nums[i];
            int l = i+1,r = nums.length-1;

            while(l < r){
                if(nums[l] + nums[r] == curTarget)
                    return target;
                else{
                    int curDiff = Math.abs(nums[l] + nums[r] - curTarget);
                    if(curDiff < diff){
                        diff = curDiff;
                        res = nums[l] + nums[r] + nums[i];
                    }

                    if(nums[l] + nums[r] < curTarget)
                        l++;
                    else
                        r--;
                }
            }
        }

        return res;
    }
}