LeetCode 16 3Sum Closest(C,C++,Java,Python)
阿新 • • 發佈:2018-12-27
Problem:
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).
Solution:
此題與15題基本類似,甚至更簡單一些,只需要比較和的結果即可,碰到和等於target的時候就直接返回吧!!!題目大意:
給一個整數陣列,找到三個數的和與給定target的值距離最短的那個和解題思路:
直接看程式碼把,沒看懂的看我的15題題解Java原始碼(用時342ms):
public class Solution { public int threeSumClosest(int[] nums, int target) { int length=nums.length,Min=Integer.MAX_VALUE; Arrays.sort(nums); for(int i=0;i<length-2;i++){ if(i>0 && nums[i]==nums[i-1])continue; int begin=i+1,end=length-1; while(begin<end){ int sum=nums[i]+nums[begin]+nums[end]; if(Math.abs(sum-target)<Math.abs(Min))Min=sum-target; if(sum==target)return target; else if(sum>target)end--; else begin++; } } return Min+target; } }
C語言原始碼(用時9ms):
int abs(int tar){ return tar>0?tar:-tar; } void quickSort(int* nums,int first,int end){ int l=first,r=end; if(first>=end)return; int temp=nums[l]; while(l<r){ while(l<r && nums[r]>=temp)r--; if(l<r)nums[l]=nums[r]; while(l<r && nums[l]<=temp)l++; if(l<r)nums[r]=nums[l]; } nums[l]=temp; quickSort(nums,first,l-1); quickSort(nums,l+1,end); } int threeSumClosest(int* nums, int numsSize, int target) { int begin,end,i,sum,Min=INT_MAX; quickSort(nums,0,numsSize-1); for(i=0;i<numsSize-2;i++){ if(i>0 && nums[i]==nums[i-1])continue; begin=i+1;end=numsSize-1; while(begin<end){ sum=nums[i]+nums[begin]+nums[end]; if(abs(sum-target)<abs(Min))Min=sum-target; if(sum==target)return target; else if(sum>target)end--; else begin++; } } return Min+target; }
C++原始碼(用時12ms):
class Solution {
public:
int threeSumClosest(vector<int>& nums, int target) {
int length=nums.size(),Min=2147483647;
quickSort(nums,0,length-1);
for(int i=0;i<length-2;i++){
if(i>0 && nums[i]==nums[i-1])continue;
int begin=i+1,end=length-1;
while(begin<end){
int sum=nums[i]+nums[begin]+nums[end];
if(abs(sum-target)<abs(Min))Min=sum-target;
if(sum==target)return target;
else if(sum>target)end--;
else begin++;
}
}
return Min+target;
}
private:
int abs(int t){
return t>0?t:-t;
}
void quickSort(vector<int>& nums,int first,int end){
int l=first,r=end,tmp;
if(first>=end)return;
tmp=nums[l];
while(l<r){
while(l<r && nums[r]>=tmp)r--;
if(l<r)nums[l]=nums[r];
while(l<r && nums[l]<=tmp)l++;
if(l<r)nums[r]=nums[l];
}
nums[l]=tmp;
quickSort(nums,first,l-1);
quickSort(nums,l+1,end);
}
};
Python原始碼(用時127ms):
class Solution:
# @param {integer[]} nums
# @param {integer} target
# @return {integer}
def threeSumClosest(self, nums, target):
length=len(nums);Min=2147483647
nums.sort()
for i in range(length-2):
if i>0 and nums[i]==nums[i-1]:continue
begin=i+1;end=length-1
while begin<end:
sum=nums[i]+nums[begin]+nums[end]
if abs(sum-target)<abs(Min):Min=sum-target
if sum==target:return target
elif sum>target:end-=1
else:begin+=1
return Min+target