1. Two Sum

Given an array of integers, return indices of the two numbers such that they add up to a specific target.

You may assume that each input would have exactly one solution.

Example:

Given nums = [2, 7, 11, 15], target = 9,Because nums[0] + nums[1] = 2 + 7 = 9,return [0, 1].

UPDATE (2016/2/13):
The return format had been changed to zero-based indices. Please read the above updated description carefully.

解法1：

用unordered_map存储vector中每个元素对应的index，遍历的时候查找是否之前已经储存过元素x，满足nums[i] + x == target。

若存在，则输出结果；若不存在，则将这个元素及其index加入unordered_map。

时间复杂度：O(n)

class Solution {public:    vector<int> twoSum(vector<int>& nums, int target) {        unordered_map<int, int> index;        for (int i = 0; i < nums.size(); i++) {            if (index.find(target - nums[i]) != index.end()) {                return vector<int>({index[target - nums[i]], i});            }            index[nums[i]] = i;        }        return vector<int>(2, -1);    }};

解法2：

先将vector中元素升序排序，再用两个游标left和right分别从vector两端遍历。

若element[left] + element[right] == target，则返回它们的index；

若element[left] + element[right] < target，则left游标向右移动；反之right游标向左移动。

由于最后需要返回元素的原始index，所以首先需要建立一个数据结构Elem记录元素对应的index和它的值。

时间复杂度：O(n log n)

class Solution {public:    struct Elem {        int index;        int val;        Elem(int i, int v) : index(i), val(v) {}    };    static bool Comp(Elem a, Elem b) {        return a.val < b.val;    }    vector<int> twoSum(vector<int>& nums, int target) {        vector<Elem> elements;        for (int i = 0; i < nums.size(); i++) {            elements.push_back(Elem(i, nums[i]));        }        sort(elements.begin(), elements.end(), Comp);        int left = 0;        int right = nums.size() - 1;        vector<int> result(2, -1);        while (left < right) {            int sum = elements[left].val + elements[right].val;            if (sum == target) {                result = vector<int>({elements[left].index, elements[right].index});                sort(result.begin(), result.end());                return result;            } else if (sum < target) {                left++;            } else {                right--;            }        }        return result;    }};