2Sum，3Sum问题

1.2sum问题

Given nums = [2, 7, 2, 11, 15], target = 9,

Because nums[0] + nums[1] = 2 + 7 = 9,

方法：排序，使用左右指针，复杂度为nlogn

class Solution {
public:
    vector<int> TwoSum(vector<int> &a, int num) {
        vector<int> result;
        sort(a.begin(), a.end());//vector中sort函数使用的是快排，时间复杂度为NlogN
        int len = a.size();
        int low = 0;
        int high = len - 1;
        while (low < high) {
            if (low == high)
                break;
            int sum = a[low] + a[high];
            if (sum < num)
                low++;
            else if (sum > num)
                high--;
            else {
                result.push_back(a[low]);
                result.push_back(a[high]);
                low++;
                high--;
            }
        }
        for (int i = 0; i < result.size(); i++) {
            cout << result[i] << endl;
        }
        return result;
    }
};

int main() {
    Solution st;
    vector<int> a = {2, 5, 3, 6, 1};
    st.TwoSum(a, 7);
}

2.3Sum问题

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

A solution set is:[ [-1, 0, 1], [-1, -1, 2]]。

方法

class Solution {
public:
    vector<vector<int>> ThreeSum(vector<int> &a) {
        vector<vector<int>> result;
        sort(a.begin(), a.end());
        int len = a.size();
        for (int i = 0; i < len; i++) {
            int target = -a[i];//对当前数取相反数
            int low = i + 1;
            int high = len - 1;
            while (low < high) {
                if (low == high)
                    break;
                int sum = a[low] + a[high];
                if (sum > target)
                    high--;
                else if (sum < target)
                    low++;
                else {
                    vector<int> tmp(3, 0);
                    tmp[0] = a[i];
                    tmp[1] = a[low];
                    tmp[2] = a[high];
                    result.push_back(tmp);
                    while (low < high && a[low] == tmp[1])//去除重复元素
                        low++;
                    while (low < high && a[low] == tmp[2])
                        high--;
                }
            }
            while (i + 1 < len && a[i + 1] == a[i])//若相邻两数相等，可跳过后面那个数
                i++;
        }
        return result;
    }
};