2014
11-19

LeetCode-3Sum Closest[数组]

3Sum Closest

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).


// LeetCode, 3Sum Closest
// 先排序，然后左右夹逼，时间复杂度O(n^2)，空间复杂度O(1)
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;
}
};


1. 换句话说，A[k/2-1]不可能大于两数组合并之后的第k小值，所以我们可以将其抛弃。
应该是，不可能小于合并后的第K小值吧

2. 如果两个序列的最后字符不匹配（即X [M-1]！= Y [N-1]）
L（X [0 .. M-1]，Y [0 .. N-1]）= MAX（L（X [0 .. M-2]，Y [0 .. N-1]），L（X [0 .. M-1]，Y [0 .. N-1]）
这里写错了吧。