题目地址: 3Sum
题目简介:给一个数组,从一个数组中挑出三个数字a,b,c满足;
Given an array nums of n integers, are there elements a, b, c in nums 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.
Example:
Given array nums = [-1, 0, 1, 2, -1, -4],
A solution set is:
[
[-1, 0, 1],
[-1, -1, 2]
]
题目解析:首先想到暴力,但是会超时。于是可以仅考虑一个数字(nums[i])的右侧数字,并根据以下规则滑动:
- 先对数组排序,从左往右滑动。假设紧挨a[i]右边的为a[l](从左边开始遍历的开头),开始为最右边的为a[r];
- 如果这三个数相加的和大于0,那么应该减小结果,则应该让a[r]向左滑动;
- 如果这三个数相加的和小于0,那么应该增大结果,则应该让a[l]向右滑动;
- 如果这三个数相加的和等于0,那么需要保存结果,为了是返回的结果中没有重复,那么需要划过排序后的数组中相同的数字。因为滑动方向为从左到右,考虑到可能存在3个0相邻的情况,那么从重复数字的最左边的数字计算,计算完成以后划到下一个数字。
C++版(从左往右滑动):
class Solution {
public:
vector<vector<int>> threeSum(vector<int>& nums) {
vector<vector<int>> ans_set;
if (nums.size() <= 2)
{
return ans_set;
}
sort(nums.begin(), nums.end());
for (int i = 0; i < nums.size() - 2 && nums[i] <= 0; i++)
{
if (i > 0 && nums[i] == nums[i - 1]) continue;
int l = i + 1, r = nums.size() - 1;
while(l < r)
{
if (nums[i] + nums[l] + nums[r] > 0) r--;
else if (nums[i] + nums[l] + nums[r] < 0) l++;
else
{
ans_set.push_back({nums[i], nums[l], nums[r]});
while(l < r && nums[l] == nums[l + 1]) l++;
while(l < r && nums[r] == nums[r - 1]) r--;
l++;
r--;
}
}
}
return ans_set;
}
};
Python(从左到右)
class Solution:
def threeSum(self, nums: List[int]) -> List[List[int]]:
ans_list = []
if (len(nums) <= 2): #如果当前的数字少于三个那么直接返回结果
return (ans_list)
nums.sort()
for i in range(0,len(nums) - 2):
if (nums[i] > 0):
break
if (i > 0 and nums[i] == nums[i - 1]):
continue
l = i + 1
r = len(nums) - 1
while (l < r):
if (nums[i] + nums[l] + nums[r] == 0):
ans_list.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
elif (nums[i] + nums[l] + nums[r] > 0):
r -=1
else:
l += 1
return (ans_list)
Java版本:
class Solution {
public List<List<Integer>> threeSum(int[] nums) {
List<List<Integer>> ans_set = new ArrayList<List<Integer>>();
if (nums.length <= 2)
{
return ans_set;
}
Arrays.sort(nums);
for (int i = 0; i < nums.length - 2 && nums[i] <= 0; i++)
{
if (i > 0 && nums[i] == nums[i - 1]) continue;
int l = i + 1, r = nums.length - 1;
while(l < r)
{
if (nums[i] + nums[l] + nums[r] > 0) r--;
else if (nums[i] + nums[l] + nums[r] < 0) l++;
else
{
ans_set.add(Arrays.asList(nums[i], nums[l], nums[r]));
while(l < r && nums[l] == nums[l + 1]) l++;
while(l < r && nums[r] == nums[r - 1]) r--;
l++;
r--;
}
}
}
return ans_set;
}
}