标签：
题目：
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).
c代码：
#define INT_MAX 2147483647L /* maximum (signed) long value */
int Partition(int a[], int low, int high){
int pivotkey = a[low];
while(low < high){
while( low < high && a[high] >= pivotkey) --high;
a[low]= a[high];
while(low < high && a[low] <= pivotkey) ++low;
a[high] = a[low];
}
a[low] = pivotkey;
return low;
}
void Qsort(int a[], int low, int high){
int pivotloc;
if(low < high){
pivotloc = Partition(a, low, high);
Qsort(a, low, pivotloc - 1);
Qsort(a, pivotloc + 1, high);
}
}
void QuikSort(int a[], int size){
Qsort(a, 0 ,size-1);
}
int threeSumClosest(int *num, int n, int target) {
int min = INT_MAX,record,i;
QuikSort(num, n);
for(i = 0; i < n - 2; i++){
int l = i + 1;
int r = n - 1;
while(l < r){
int sum = num[i] + num[l] + num[r];
if(sum == target){
return sum;
}else if(sum < target){
if(target - sum < min){
record = sum;
min = target - sum;
}
l++;
}else{
if(sum - target < min){
record = sum;
min = sum - target;
}
r--;
}
}
while(i+1<n-2 && num[i+1] == num[i])i++;
}
return record;
}
标签：
原文地址：http://www.cnblogs.com/zhutianpeng/p/4304577.html
