Problem Description
Given an array S with n elements, are there elements a, b, and c in S satisfying a+b+c=0? Find all triples in the array S that meet the conditions.
Note:
The elements in the triples (a, b, c) must be arranged in non-descending order. (That is, a≤b≤c) The
solution set cannot contain repeated triples.
Input description:
Input an array
Output description:
output an array set with the sum of three numbers being 0
Example
Example 1
Enter
[-2,0,1,1,2]
Output
[[-2,0,2],[-2,1,1]]
Solutions
analysis
- Using the method of three pointer subscripts, the subscript i can be used as the subscript of the traversal array, left is the left subscript greater than i, and right is the right subscript less than len-1. The movement is determined by judging the size relationship between the three elements and 0 Left subscript or right subscript.
method
- A three-pointer subscript method is used.
Code
// 思路1
public class Solution {
public ArrayList<ArrayList<Integer>> threeSum(int[] num) {
Arrays.sort(num);
ArrayList<ArrayList<Integer>> res = new ArrayList<>();
ArrayList<Integer> t;
int len = num.length;
for (int i = 0; i < len - 2; i++) {
//len-3,len-2,len-1
if (i == 0 || num[i] != num[i - 1]) {
int left = i + 1; // 左下标
int right = len - 1; // 右下标
while (left < right) {
// 如果三个数字和大于0，右下标左移动
while (left < right && (num[i] + num[left] + num[right]) > 0) {
right--;
}
// 符合条件，添加元素
if (left < right && (num[i] + num[left] + num[right]) == 0) {
t = new ArrayList<>();
t.add(num[i]);
t.add(num[left]);
t.add(num[right]);
res.add(t);
// 如果最小元素相同，左下标右移动
while (left < right && num[left] == t.get(1)) {
left++;
}
} else {
// 如果小于0，则左下标右移动
left++;
}
}
}
}
return res;
}
}
Time complexity analysis: The
first time: n-2 The
second time: n-3
The n-
2th time: 1 and = (n-2) + (n-3) +… + 1 = 1/2 * n^ + 3/2 * n + 1 = O(N^)
The time complexity of java quick sort: O(n * logn)
So the time complexity here is: O(N^)
Space complexity analysis:
extra space t is used here, but t only stores 3 numbers, so the space complexity is: O(1)
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