題目鏈接:https://leetcode.com/problems/3sum/
題目: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)
題意:給定一個數組,找出所有和爲0的三個數,添加到ArrayList<Interger>中。要求這三個數按非降序排列,並且ArrayList不能重複。
示例代碼:
public class Solution
{
public List<List<Integer>> threeSum(int[] nums)
{
List<List<Integer>> result=new ArrayList<List<Integer>>();
int length=nums.length;
if(nums==null||length<3)
return result;
//對目標數組按照升序排序
Arrays.sort(nums);
for(int i=0;i<length-2;i++)
{
//略過相同的數
if(i!=0&&nums[i]==nums[i-1])
{
continue;
}
int left=i+1;
int right=length-1;
//在nums[i]的右側尋找兩個數,值得它們爲0
while(left<right)
{
int a1=nums[left];
int a2=nums[right];
int sum=a1+a2+nums[i];
if (sum==0)
{
ArrayList<Integer> list=new ArrayList<Integer>();
list.add(nums[i]);
list.add(a1);
list.add(a2);
result.add(list);
left++;
right--;
while (left < right && nums[left] == nums[left - 1])
{
//左側略過相同的結果
left++;
}
while (left < right && nums[right] == nums[right + 1])
{
//右側略過相同的結果
right--;
}
}
else if(sum>0) //sum>0,說明右側的數大了,應該往左邊移動
{
right--;
}
else //反之要往右邊移動
{
left++;
}
}
}
return result;
}
}