蜗牛慢慢爬 LeetCode 15. 3Sum [Difficulty: Medium]
题目
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: 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]
]
翻译
给定一个数组S,它包含n个整数,它是否存在3个元素a,b,c,满足a+b+c=0?找出所有满足条件的元素数组。
提示:a,b,c三个元素必须是升序排列(也就是满足a ≤ b ≤ c),最终的结果不能包含重复的元素数组。例如给定S为{-1 0 1 2 -1 -4},返回结果是(-1, 0, 1)和(-1, -1, 2)。
Hints
Related Topics: Array, Two Pointers
最容易想到的应该就是三重循环搞定 为了减少时间复杂度 自己原本是想到将最后一重循环用哈希表替换然后达到O(n^2)的效果 其中先排序可以减少很多麻烦
不过tag里面是有Two Pointers的 应该想到排序之后将二三重循环用两个指针夹逼 discuss中也是这么做的
代码
Java
class Solution {
public List<List<Integer>> threeSum(int[] nums) {
Array.sort(nums);
List<List<Integer>> res = new ArrayList<>();
for(int i=0;i<nums.length-2;i++){
int lo = i+1; int hi = num.length-1; int sum = -nums[i];
while(lo<hi){
if(nums[lo]+nums[hi]==sum){
res.add(Arrays.asList(nums[i],nums[lo],nums[hi]));
while(lo<hi && nums[lo]==nums[lo+1]) lo++;
while(lo<hi && nums[hi]==nums[hi-1]) hi--;
lo++;hi--;
}else if(nums[lo]+nums[hi]<sum){
lo++;
}else{
hi--;
}
}
}
return res;
}
}
Python
class Solution(object):
def threeSum(self, nums):
"""
:type nums: List[int]
:rtype: List[List[int]]
"""
nums.sort()
mymap = {}
result = []
for i in range(len(nums)):
if nums[i] not in mymap:
l = []
l.append(i)
mymap[nums[i]] = l
else:
mymap[nums[i]].append(i)
for i in range(0, len(nums)-2):
if nums[i]>0:
break
if i>0 and nums[i]==nums[i-1]:
continue
for j in range(i+1, len(nums)-1):
if j>i+1 and nums[j]==nums[j-1]:
continue
target = -nums[i]-nums[j]
if target<nums[j]:
break
mylist = []
if target in mymap:
mylist = mymap[target]
else:
continue
for integer in mylist:
if integer!=i and integer!=j:
res = [nums[i],nums[j],nums[integer]]
result.append(res)
break
return result
//better solution from discuss
class Solution(object):
def threeSum(self, nums):
res = []
nums.sort()
for i in xrange(len(nums)-2):
if i > 0 and nums[i] == nums[i-1]:
continue
l, r = i+1, len(nums)-1
while l < r:
s = nums[i] + nums[l] + nums[r]
if s < 0:
l +=1
elif s > 0:
r -= 1
else:
res.append((nums[i], nums[l], nums[r]))
while l < r and nums[l] == nums[l+1]:
l += 1
while l < r and nums[r] == nums[r-1]:
r -= 1
l += 1; r -= 1
return res
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