leetcode | 3Sum

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问题描述：点击打开链接 leetcode

Given an array S of n integers, are there elements a, b, c in S such that a + b + c = 0? Find all unique triplets in the array which gives the sum of zero.

Note:

• Elements in a triplet (a,b,c) must be in non-descending order. (ie, a ≤ b ≤ c)
• The solution set must not contain duplicate triplets.

    For example, given array S = {-1 0 1 2 -1 -4},

    A solution set is:
    (-1, 0, 1)
    (-1, -1, 2)

解析：

首先排序，然后左右夹逼，复杂度O(n^2) -- 从程序中可以看出

此方法可扩展到 k-sum （即求k个数的和等于target）,先排序，然后做 k-2 次循环，在最内层循环中左右夹逼；时间复杂度为\[\max (O(nlogn),O({n^{k - 1}}))\]

（排序O(nlogn), 循环\[O({n^{k - 1}})\]）

#include <iostream>
#include <vector>
#include <algorithm>

using namespace std;
class Solution {
public:
	vector<vector<int>> threeSum(vector<int>& nums) {
		vector< vector<int> > result;
		if (nums.size() < 3)
			return result;
		sort(nums.begin(), nums.end());
		int target = 0;
		for (int i = 0; i < nums.size()-2; i++) {
			int j = i+1;
			int k = nums.size() - 1;
			if (i > 0 && nums[i] == nums[i-1])
				continue;
			while (j < k) {
				int sum = nums[i] + nums[j] + nums [k];
				if (sum < target) {
					j++;
					while (j < k && nums[j] == nums[j-1])
						j++;
				} else if (sum > target) {
					k--;
					while (j < k && nums[k] == nums[k+1])
						k--;
				} else {
					int t[] = {nums[i], nums[j], nums[k]};
					vector<int> temp(t, t+3);
					result.push_back(temp);
					cout << nums[i] << " "<< nums[j] << " " << nums[k] << endl;
					j++;
					k--;
					while (j < k && nums[j] == nums[j-1] && nums[k] == nums[k+1])//只有两个都等时，需要改变一个
						j++;
				}
			}
		}
		return result;
	}

};
int main()
{
	int test1[] = {-1, 0, 1, 2, -1, -4};
	int test2[] = {-1, 0, 1};
	int test4[] = {-1, 0};
	int test3[] = {2, 3, -2, 0, -10, 4, 6};

	vector<int> nums(test1, test1+6);
	vector<int> nums2(test2, test2+3);
	vector<int> nums3(test3, test3+7);
	vector<int> nums4(test4, test4+2);
	Solution sol;
	sol.threeSum(nums);
	sol.threeSum(nums2);
	sol.threeSum(nums3);
	sol.threeSum(nums4);
}

leetcode | 3Sum

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原文地址：http://blog.csdn.net/quzhongxin/article/details/45888145