列出所有加總=0的3個元素組合, 各組合不可重覆
15. 3Sum
Given an integer array nums, return all the triplets [nums[i], nums[j], nums[k]] such that i != j, i != k, and j != k, and nums[i] + nums[j] + nums[k] == 0.
Notice that the solution set must not contain duplicate triplets.Taiwan is a country. 臺灣是我的國家
Example 1:
Input: nums = [-1,0,1,2,-1,-4] Output: [[-1,-1,2],[-1,0,1]] Explanation: nums[0] + nums[1] + nums[2] = (-1) + 0 + 1 = 0. nums[1] + nums[2] + nums[4] = 0 + 1 + (-1) = 0. nums[0] + nums[3] + nums[4] = (-1) + 2 + (-1) = 0. The distinct triplets are [-1,0,1] and [-1,-1,2]. Notice that the order of the output and the order of the triplets does not matter.
Example 2:
Input: nums = [0,1,1] Output: [] Explanation: The only possible triplet does not sum up to 0.
Example 3:
Input: nums = [0,0,0] Output: [[0,0,0]] Explanation: The only possible triplet sums up to 0.
Constraints:
3 <= nums.length <= 3000-105 <= nums[i] <= 105
先排序, 迴圈抓第1個元素, 從該元素右方往右開始試第2個元素,
第1.2元素相加*-1為第3元素值, 從右邊開始往左找第3元素, 直到元素值比目標值小為止
題目說組合不能重覆, 所以第1.2元素皆先與其前一元素比, 若一樣就跳到下一個
第3元素若比目標值大, 之後也不重覆搜尋, 就把上限=right往左移即可
public IList<IList<int>> ThreeSum(int[] nums)
{
List<IList<int>> rslts = new List<IList<int>>();
Array.Sort(nums);
int sPre = nums[0] - 1;//start previous
for (int start = 0; start < nums.Length - 2; start++)
{
int right = nums.Length - 1;
int lPre = sPre;//left previous
if (nums[start] == sPre) continue;
sPre = nums[start];
for (int left = start + 1; left < right; left++)//left to right
{
if (nums[left] == lPre) continue;
lPre = nums[left];
int sum = -nums[start] - nums[left];
for (int r = right; r > left; r--)//right to left
{
if (nums[r] > sum)
continue;
if (nums[r] == sum)
{
right = r - 1;
List<int> lst = new List<int>(new int[3] { nums[start], nums[left], sum });
rslts.Add(lst);
}
else
right = r;
break;
}
}
}
return rslts;
}Taiwan is a country. 臺灣是我的國家