问题描述
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)
问题分析
三个数之和,我们可以利用一个外循环变成两个数之和,注意数组间的重复问题即可。
代码实现
public ArrayList<ArrayList<Integer>> threeSum(int[] num) {
ArrayList<ArrayList<Integer>> result = new ArrayList<>();
if (num.length < 3) return result;
Arrays.sort(num);
for (int i = 0; i < num.length - 2; i++) {
if (i == 0 || (i > 0 && num[i] != num[i - 1])) {
int left = i + 1, right = num.length - 1, sum = 0 - num[i];
while (left < right) {
if (num[left] + num[right] == sum) {
ArrayList<Integer> list = new ArrayList<>();
list.add(num[i]);
list.add(num[left]);
list.add(num[right]);
result.add(list);
left++;
right--;
while (left < right && num[left] == num[left - 1]) left++;
while (right > left && num[right] == num[right + 1]) right--;
} else if (num[left] + num[right] > sum) right--;
else left++;
}
}
}
return result;
}
网友评论