LeetCode:16. 3Sum Closest
题目:
LeetCode: 16. 3Sum Closest
描述:
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.
找出数组中三个元素和为目标数的序列集合。
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:
本题思路和15. 3Sum基本上一致的。
思路:
1、由于是找出和为target的元素,首先对vector进行排序,有利于进一步处理。
2、三个数字和首先想到的是固定一个数值,另两个数字作为游标遍历vector。例如:[-1, 0, 1, 2, -1, -4]中,排序后为[ -4, -1, -1, 0, 1, 2] 固定最小的首个数字为k1=-4,令k2从 -1 向右移动,k3 = 2向左移动。直至 k3游标移到k2游标位置,
3、根据k2 + k3 + k1 ? target 的关系,来决定游标移动方向,k1固定,当k2 + k3 + k1 > 0时,k3 左移(减小三个数和),反之k2右移。当相等时,存入数组,并令k2 k3同时移动,便面出现多组重复集合的情况。
代码:
int threeSumClosest(vector<int>& vecNum, int nTarget) {
if (vecNum.size() < 3)
{
return INT_MAX;
}
sort(vecNum.begin(), vecNum.end());
int nDiff = INT_MAX;
for (vector<int>::iterator iterBg = vecNum.begin(); iterBg < vecNum.end() - 2; iterBg++)
{
if (iterBg > vecNum.begin() && *(iterBg - 1) == *iterBg)
{
continue;
}
vector<int>::iterator iterL = iterBg + 1;
vector<int>::iterator iterR = vecNum.end() - 1;
while (iterL < iterR)
{
int nSum = *iterBg + *iterL + *iterR;
if (nTarget > nSum)
{
iterL++;
while (*iterL == *(iterL - 1) && iterL < iterR)
{
iterL++;
}
}
else if (nTarget < nSum)
{
iterR--;
while (*iterR == *(iterR + 1) && iterL < iterR)
{
iterR--;
}
}
else
{
return nTarget;
}
int nTemp = abs(nTarget - nSum);
if (abs(nDiff) > nTemp)
{
nDiff = nTarget - nSum;
}
}
}
return nTarget - nDiff;
}
备注:
可以参考15. 3Sum.