我正在尝试解决一个问题,
Given an array S of n integers, find three integers in S such that the sum is closest to a given number, target. Return the sum of the three integers. You may assume that each input would have exactly one solution.
For example, given array S = {-1 2 1 -4}, and target = 1. The sum that is closest to the target is 2. (-1 + 2 + 1 = 2).
我的解决方案是:从数组中取出一个数字 (number_1),将目标设置为 target - 该数字并找到另外两个最接近新目标的数字。这样:number_1 + number_2 + number_3 将最接近,因为 number_2 + number_3 将最接近目标 - number_1。
我在 https://leetcode.com/problems/3sum-closest/description/ 尝试了我的解决方案.
我的解决方案是:
def threeSumClosest(nums, target):
"""
:type nums: List[int]
:type target: int
:rtype: int
"""
closest_sum = nums[0] + nums[1] + nums[2]
for i in range(len(nums)):
# Create temp array excluding a number
if i!=len(nums)-1:
temp = nums[:i] + nums[i+1:]
else:
temp = nums[:len(nums)-1]
# Sort the temp array and set new target to target - the excluded number
temp = sorted(temp)
l, r = 0, len(temp) -1
t = target - nums[i]
while(l<r):
if temp[l] + temp[r] == t:
return target
elif temp[l] + temp[r] > t:
if abs(temp[l] + temp[r] + nums[i] - target) < abs(closest_sum - target):
closest_sum = temp[l] + temp[r] + nums[i]
r = r - 1
else:
if abs(temp[l] + temp[r] + nums[i] - target) < abs(closest_sum - target):
closest_sum = temp[l] + temp[r] + nums[i]
l = l + 1
return closest_sum
它通过了 125 个测试用例中的 80 个,因此解决方案逻辑对我来说看起来足够好。
它失败了:
Input:
[0,2,1,-3]
1
Output:
3
Expected:
0
无法理解为什么会失败以及如何让我的逻辑保持一致。
感谢您的帮助。
最佳答案
你有几个错误,第一个是愚蠢的,你在 return closest
中有一个额外的缩进,第二个没有检查更新 closest
在第三个 if 语句.
此代码已被接受:
class Solution(object):
def threeSumClosest(self, nums, target):
"""
:type nums: List[int]
:type target: int
:rtype: int
"""
closest = nums[0] + nums[1] + nums[2]
#if len(nums)==3:
# return closest
for i in range(len(nums)):
if i!=len(nums)-1:
temp = nums[:i] + nums[i+1:]
else:
temp = nums[:len(nums)-1]
temp = sorted(temp)
l, r = 0, len(temp) -1
t = target - nums[i]
while(l < r):
if abs(temp[l] + temp[r] + nums[i] - target) < abs(closest - target):
closest = temp[l] + temp[r] + nums[i]
if temp[l] + temp[r] == t:
return target
elif temp[l] + temp[r] > t:
r = r - 1
else:
l = l + 1
return closest
这是一个公认的 C++ 解决方案,运行时间为 O(n^2)
:
class Solution {
public:
int threeSumClosest(vector<int>& nums, int target) {
sort(nums.begin(), nums.end());
int ans = nums[0] + nums[1] + nums[2];
for(int i = 0; i < nums.size() - 2; i++) {
int l = i + 1, r = nums.size() - 1;
while (l < r) {
if(abs(nums[i] + nums[l] + nums[r] - target) < abs(target - ans)) {
ans = nums[i] + nums[l] + nums[r];
}
if(nums[r] + nums[l] > target - nums[i]) r = r - 1;
else l = l + 1;
}
}
return ans;
}
};
关于python - 找出总和最接近给定数字的三个数字,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/49589941/