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].

思路：先將輸入的array排序，再用前後夾擊的方式，取得答案。如果 兩數相加 < 目標，代表左邊的index要往下一個移動。如果 兩數相加 > 目標，代表右邊的index要往前一個移動。typedef struct Number{int index;int value;}Number;bool compare(const Number & num1,const Number & num2){return num1.value < num2.value;}class Solution {public:vector<int> twoSum(vector<int>& nums, int target){

<span style="white-space:pre">	</span>// 因為排序完 index就會錯亂，先用Number記起來
vector<Number> my_nums;
Number number;
for(int index = 0; index < nums.size(); index++) {
number.value = nums[index];
number.index = index;
my_nums.push_back(number);
}
 

<span style="white-space:pre">	</span>//排序
sort(my_nums.begin(), my_nums.end(), compare);

vector<int> result;
int begin = 0;
int tail = my_nums.size() - 1;
 

<span style="white-space:pre">	</span>//開始夾擊

for (int index = 0; index < nums.size(); index++) {

if(my_nums[begin].value + my_nums[tail].value < target) {
begin++;
continue;
}

if(my_nums[begin].value + my_nums[tail].value > target) {
tail--;
continue;
}

<pre name="code" class="cpp" style="color: rgb(51, 51, 51); font-size: 14px;">            if(my_nums[begin].value + my_nums[tail].value == target) {if (my_nums[begin].index > my_nums[tail].index) {result.push_back(my_nums[tail].index);result.push_back(my_nums[begin].index);} else {result.push_back(my_nums[begin].index);result.push_back(my_nums[tail].index);}return result;}

} return result; }};