動機
在相等時不用煩惱該縮哪邊的two sum
Problem
You are given an integer array nums
and an integer k
.
In one operation, you can pick two numbers from the array whose sum equals k
and remove them from the array.
Return the maximum number of operations you can perform on the array.
Example 1:
Input: nums = [1,2,3,4], k = 5Output: 2Explanation: Starting with nums = [1,2,3,4]:- Remove numbers 1 and 4, then nums = [2,3]- Remove numbers 2 and 3, then nums = []There are no more pairs that sum up to 5, hence a total of 2 operations.
Example 2:
Input: nums = [3,1,3,4,3], k = 6Output: 1Explanation: Starting with nums = [3,1,3,4,3]:- Remove the first two 3's, then nums = [1,4,3]There are no more pairs that sum up to 6, hence a total of 1 operation.
Constraints:
1 <= nums.length <= 105
1 <= nums[i] <= 109
1 <= k <= 109
Sol
class Solution:
def maxOperations(self, nums: List[int], k: int) -> int:
nums.sort()
a,b,ret=0,len(nums)-1,0
while a<b:
if nums[a]+nums[b] == k:
ret,a,b = ret+1, a+1, b-1
elif nums[a]+nums[b] > k:
b -= 1
else:
a += 1
return ret