我的解法是利用2Sum相同的算法:
1. 对原数组排序
2. 设置首尾两个指针。当两数的和大于target时,尾指针向左移一位;当两数的和小于target时,尾指针向右移一位。
对于4Sum的解法,因为是四个数字,利用两个循环数组确定前两个数字,剩下的两个数字和target便可以使用2Sum的算法完成。但是算法的运行时间很长。
代码如下
class Solution:
def fourSum(self, nums, target):
ans = []
nums.sort()
for i in range(len(nums)-2):
if i > 0 and nums[i] == nums[i-1]:
continue
for j in range(i+1,len(nums)-2):
if j > i+1 and nums[j] == nums[j-1]:
continue
b, e = j + 1, len(nums) - 1
while b < e:
diff = target - (nums[i] + nums[j] + nums[b] + nums[e])
if diff > 0:
b += 1
elif diff < 0:
e -= 1
else:
ans.append([nums[i], nums[j], nums[b], nums[e]])
while b < e and nums[b] == nums[b + 1]:
b += 1
while b < e and nums[e] == nums[e - 1]:
e -= 1
b += 1
e -= 1
return ans
在discuss区中的python最快代码如下:
def fourSum(self, nums, target):
def findNsum(nums, target, N, result, results):
if len(nums) < N or N < 2 or target < nums[0]*N or target > nums[-1]*N: # early termination
return
if N == 2: # two pointers solve sorted 2-sum problem
l,r = 0,len(nums)-1
while l < r:
s = nums[l] + nums[r]
if s == target:
results.append(result + [nums[l], nums[r]])
l += 1
while l < r and nums[l] == nums[l-1]:
l += 1
elif s < target:
l += 1
else:
r -= 1
else: # recursively reduce N
for i in range(len(nums)-N+1):
if i == 0 or (i > 0 and nums[i-1] != nums[i]):
findNsum(nums[i+1:], target-nums[i], N-1, result+[nums[i]], results)
results = []
findNsum(sorted(nums), target, 4, [], results)
return results
这个代码核心也是使用2Sum的算法,使用递归可解任意个数的NSum问题。但是速度快到只要120ms,问题在于少做了很多次对于重复数字的判断。