算法分析与设计——LeetCode:16. 3Sum Closest

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).

class Solution {
public:
int threeSumClosest(vector<int>& nums, int target) {
return closest;
}
};

思路

如果遍历所有情况，复杂度将为O(n^3)。为了降低其复杂度，先按大小将序列进行排序，调用sort函数，复杂度为O(nlog2n)，然后如下图

横条代表排序后的序列，first从头到尾遍历一遍，每一轮中，second指向first+1，third指向序列最后一个数，若和大于target，则将third前移一位，使其变小，若和小于target，则将second后移一位，这样最后就可以找到与target最接近的和，算法复杂度为O(n^2)。

代码

class Solution {
public:
int threeSumClosest(vector<int>& nums, int target) {
sort(nums.begin(), nums.end());
int first, second, third;
int size = nums.size();
int closest;
int min;
closest = nums[0]+nums[1]+nums[2];
min = abs(target-closest);
for (first = 0; first < size-2; first++) {
int sum;
second = first+1;
third = size-1;
while (second != third) {
sum = nums[first]+nums[second]+nums[third];
if (sum == target) {
return target;
}
int d = abs(target-sum);
if (d < min) {
min = d;
closest = sum;
}
if (sum < target) {
second++;
} else {
third--;
}
}
}
return closest;
}
};