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LeetCode_15:三数之和(Python+ Java)

热度:11   发布时间:2024-02-10 22:56:41.0
  • 题目描述:

??????给你一个包含 n 个整数的数组 nums,判断 nums 中是否存在三个元素 a,b,c ,使得 a + b + c = 0 ?请你找出所有满足条件且不

重复的三元组。

注意:答案中不可以包含重复的三元组。

 

  • 示例:

 

给定数组 nums = [-1, 0, 1, 2, -1, -4],满足要求的三元组集合为:
[[-1, 0, 1],[-1, -1, 2]
]

 

  • 解法一:???????参考链接:https://leetcode-cn.com/problems/3sum/solution/pai-xu-shuang-zhi-zhen-zhu-xing-jie-shi-python3-by/
#Python
class Solution(object):def threeSum(self, nums):n = len(nums)res = []if not nums and n < 3:return resnums.sort()for i in range(0, n):   #外层循环,定位if nums[i] > 0: breakif i > 0 and nums[i] == nums[i - 1]: continueL = i + 1R = n - 1while L < R:        #内层循环, 定值temp = nums[i] + nums[L] + nums[R]if temp == 0: res.append([nums[i], nums[L], nums[R]])while (L < R) and (nums[L] == nums[L + 1]):L += 1while (L < R) and (nums[R] == nums[R - 1]):R -= 1L=L+1R=R-1elif(temp>0):R=R-1else:L=L+1return res
//Java
class Solution {public List<List<Integer>> threeSum(int[] nums) {int len = nums.length;List<List<Integer>> res = new ArrayList<>();Arrays.sort(nums);for (int i = 0; i < len; i++) {if(nums[i] > 0) return res;if(i > 0 && nums[i] == nums[i-1]) continue;int L = i + 1;int R = len - 1;while(L <R){int temp = nums[i] + nums[L] + nums[R];if(temp == 0){List<Integer> list = new ArrayList<>();list.add(nums[i]);list.add(nums[L]);list.add(nums[R]);res.add(list);while(L < R && nums[L] == nums[L + 1]) ++L;while(L < R && nums[R] == nums[R - 1]) --R;++L;--R;}else if(temp < 0) ++L;else if(temp > 0) --R;}}return res;}
}

 

 

 

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