【JAVA、C++】LeetCode 016 3Sum Closest

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).

同上题，没上一题难，比葫芦画瓢即可

JAVA实现：

static public int threeSumClosest(int[] nums, int target) {
		int result = nums[0] + nums[1] + nums[2];
		Arrays.sort(nums);
		for (int i = 0; i < nums.length - 2; i++) {
			int j = i + 1, k = nums.length - 1;
			while (j < k) {
				if (Math.abs(nums[i] + nums[j] + nums[k] - target) < Math.abs(result - target))
					result = nums[i] + nums[j] + nums[k];
				if (nums[i] + nums[j] + nums[k] < target)
					j++;
				else if (nums[i] + nums[j] + nums[k] > target)
					k--;
				else
					return target;
				//不加while循环亦可通过测试，但是时间会偏长一些
				while (i < k && nums[i] == nums[i + 1])
					i++; 
			}
		}
		return result;
	}

 C++:

 1 class Solution {
 2 public:
 3     int threeSumClosest(vector<int>& nums, int target) {
 4         int result = nums[0] + nums[1] + nums[2];
 5         sort(nums.begin(),nums.end());
 6         for (int i = 0; i < nums.size() - 2; i++) {
 7             int j = i + 1, k = nums.size() - 1;
 8             while (j < k) {
 9                 if (abs(nums[i] + nums[j] + nums[k] - target) < abs(result - target))
10                     result = nums[i] + nums[j] + nums[k];
11                 if (nums[i] + nums[j] + nums[k] < target)
12                     j++;
13                 else if (nums[i] + nums[j] + nums[k] > target)
14                     k--;
15                 else
16                     return target;
17                 while (i < k && nums[i] == nums[i + 1])
18                     i++;
19             }
20         }
21         return result;
22     }
23 };

 

posted @ 2015-04-29 20:14  TonyLuis  阅读(269)  评论(0)    收藏  举报