
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).
分析:
先排序,然后左右夹逼,复杂度O(n^2)
//方法一:先排序,然后左右夹逼,复杂度O(n^2)
class Solution {
public:
int threeSumClosest(vector<int>& nums, int target) {
if(nums.size() == 0)
return 0;
if(nums.size() == 1)
return nums[0];
if(nums.size() == 2)
return nums[0] + nums[1];
if(nums.size() == 3)
return nums[0] + nums[1] + nums[2];
sort(nums.begin(), nums.end());
int result = *nums.begin() + *(nums.begin() + 1) + *(nums.end() - 1);
for(vector<int>::iterator i = nums.begin(); i < nums.end() - 2; i++)
{
if(i > nums.begin() && *i == *(i-1))
continue;
vector<int>::iterator j = i + 1;
vector<int>::iterator k = nums.end() - 1;
while(j < k)
{
if(*i + *j + *k < target)
{
if(abs(*i + *j + *k - target) < abs(result - target))
result = *i + *j + *k;
j++;
while(*j == *(j-1) && j < k)
j++;
}
else if(*i + *j + *k > target)
{
if(abs(*i + *j + *k - target) < abs(result - target))
result = *i + *j + *k;
k--;
while(*k == *(k+1) && j < k)
k--;
}
else
return target;
}
}
return result;
}
};
//方法二:参考
class Solution {
public:
int threeSumClosest(vector<int>& num, int target) {
int result = 0;
int min_gap = INT_MAX;
sort(num.begin(), num.end());
for (auto a = num.begin(); a != prev(num.end(), 2); ++a)
{
auto b = next(a);
auto c = prev(num.end());
while (b < c)
{
const int sum = *a + *b + *c;
const int gap = abs(sum - target);
if (gap < min_gap)
{
result = sum;
min_gap = gap;
}
if (sum < target)
++b;
else
--c;
}
}
return result;
}
};
原文地址:http://blog.csdn.net/keyyuanxin/article/details/46581145