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LeetCode 算法题 15. 3Sum

分类:algorithm| 发布时间:2016-07-03 01:04:00


题目

Given an array S of n integers, are there elements a, b, c in S such that a + b + c = 0? Find all unique triplets in the array which gives the sum of zero.

Note: The solution set must not contain duplicate triplets.

For example, given array S = [-1, 0, 1, 2, -1, -4],

A solution set is:

[
    [-1, 0, 1],
    [-1, -1, 2]
]

题意

给出一个数组找出任意三个元素相加等于 0。 找出所有情况,注意结果集不能有重复的元组。

解法 1

先将给出的数组排序。通过 O(n3) 的算法找出所有匹配的结果, 注意处理重复元素的问题。

class Solution {
public:
    vector<vector<int>> threeSum(vector<int>& nums) {
        vector<vector<int>> ret;
        const int size = nums.size();
        if (size < 3) {
            return ret;
        }

        sort(nums.begin(), nums.end());
        int expect;
        for (int i = 0; i < size; ++i) {
            for (int j = i + 1; j < size - 1; ++j) {
                expect = -nums[i] - nums[j];
                for (int k = j + 1; k < size; ++k) {
                    if (nums[k] == expect) {
                        ret.push_back({nums[i], nums[j], nums[k]});
                        while (k < size - 1 && nums[k] == nums[k + 1]) {
                            ++k;
                        }
                    }
                }

                while (j < size - 1 && nums[j] == nums[j + 1]) {
                    ++j;
                }
            }

            while (i < size && nums[i] == nums[i + 1]) {
                ++i;
            }
        }

        return ret;
    }
};

解法 2

上述算法的复杂度太高了,其实稍微修改下里面的两个循环就能简化成一个。

class Solution {
public:
    vector<vector<int> > threeSum(vector<int> &num) {
        vector<vector<int> > res;
        sort(num.begin(), num.end());
        for (int i = 0; i < num.size(); i++) {
            int target = -num[i];
            int front = i + 1;
            int back = num.size() - 1;

            while (front < back) {
                int sum = num[front] + num[back];
                // Finding answer which start from number num[i]
                if (sum < target) {
                    front++;
                } else if (sum > target) {
                    back--;
                } else {
                    vector<int> triplet(3, 0);
                    triplet[0] = num[i];
                    triplet[1] = num[front];
                    triplet[2] = num[back];
                    res.push_back(triplet);

                    // Processing duplicates of Number 2
                    // Rolling the front pointer to the next different number forwards
                    while (front < back && num[front] == triplet[1]) front++;

                    // Processing duplicates of Number 3
                    // Rolling the back pointer to the next different number backwards
                    while (front < back && num[back] == triplet[2]) back--;
                }
            }

            // Processing duplicates of Number 1
            while (i + 1 < num.size() && num[i + 1] == num[i]) {
                i++;
            }
        }

        return res;
    }
};