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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).

标签: Array Two Pointers
分析

先排序,然后左右夹逼,复杂度 $O(n^2)$。

代码1

// 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。