3 Sum Closest

问题描述

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)

解决思路

1. 搜索（超时）；

2. 排序 + 二分， 时间复杂度为O(n^2).

程序

public class ThreeSumClosest {
	// dfs
	private int diff = 0;
	private int abs = Integer.MAX_VALUE;

	public int threeSumClosest(int[] nums, int target) {
		if (nums == null || nums.length < 3) {
			return -1;
		}

		List<Integer> sol = new ArrayList<Integer>();
		boolean[] used = new boolean[nums.length];
		helper(nums, target, sol, used, 0);

		return target - diff;
	}

	private void helper(int[] nums, int target, List<Integer> sol,
			boolean[] used, int begin) {
		if (sol.size() == 3) {
			if (Math.abs(target) < abs) {
				abs = Math.abs(target);
				diff = target;
			}
			return;
		}
		for (int i = begin; i < nums.length; i++) {
			if (used[i]) {
				continue;
			}
			used[i] = true;
			sol.add(nums[i]);
			helper(nums, target - nums[i], sol, used, begin + 1);
			used[i] = false;
			sol.remove(sol.size() - 1);
		}
	}

	// sort + bs
	public int threeSumClosest(int[] nums, int target) {
		if (nums == null || nums.length < 3) {
			return -1;
		}

		Arrays.sort(nums);
		int diff = 0;
		int abs = Integer.MAX_VALUE;

		for (int i = 0; i < nums.length - 2; i++) {
			int begin = i + 1;
			int end = nums.length - 1;
			while (begin < end) {
				int sum = nums[i] + nums[begin] + nums[end];
				if (sum == target) {
					return sum;
				}
				if (sum < target) {
					++begin;
				} else {
					--end;
				}
				if (Math.abs(sum - target) < abs) {
					abs = Math.abs(sum - target);
					diff = sum - target;
				}
			}
		}

		return target + diff;
	}
}