一起刷LeetCode1-Two Sum

　　感觉有必要重新刷刷题了，为以后找工作做做准备，选择LeetCode+topcoder上的Data Science Tutorials，

争取每天晚上10:00开始刷一道，复习一下相关知识点。

-------------------------------------------------------分割线啦-----------------------------------------------------------------------

Two Sum

Given an array of integers, find two numbers such that they add up to a specific target number.

The function twoSum should return indices of the two numbers such that they add up to the target,

where index1 must be less than index2. Please note that your returned answers (both index1 and index2) are not zero-based.

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

Input: numbers={2, 7, 11, 15}, target=9
Output: index1=1, index2=2

【题意】：这题意思就是给定一些数，再给你一个目标数字，让你在给定的那些数字中找出两个数，这两个数的和等于目标数字。注意给定的这些数字是无序的。

【心路历程】：看到这题，简单想了想，就知道O(n*2)的遍历一定会超时。于是进一步的思考，感觉特别像hash，可是懒得实现hash。。。

就开始想想还有没有别的方法。感觉这种无序的数字，排序后都会比较好处理一点。就开始往排序上想，发现排好序后确实比较好处理。

我们用两个下标，一个指向开始beg，一个指向末尾end，我们去比较target减去beg指向的数字，与end指向的数字。如果target-beg == end ，就找到了；

如果 target-beg > end ，就beg++；如果target-beg < end ，就end--。这样的话，时间复杂度是O(n*logn)。我感觉差不多可以接受，就交了，ac。

为了学习知识，ac不是目的，我看了一下官方解法：

O(n2) runtime, O(1) space – Brute force:

The brute force approach is simple. Loop through each element x and find if there is another value that equals to target – x.

As finding another value requires looping through the rest of array, its runtime complexity is O(n2).

O(n) runtime, O(n) space – Hash table:

We could reduce the runtime complexity of looking up a value to O(1) using a hash map that maps a value to its index.

- -，最佳答案就是Hash，O(n)。

---------------------------------------------------------又是分割线啦----------------------------------------------------------------------

附上代码：

(1)这个是O(n*logn)的排序+首尾下标：

 /**
  * Note: The returned array must be malloced, assume caller calls free().
  */
 struct node {
         int num;
         int local;
 };

 int cmp(const void * a,const void *b) {
     struct node * aa = (struct node *)a;
     struct node * bb = (struct node *)b;
     if(aa->num == bb->num) {
         return aa->local - bb->local;
     }else{
         return aa->num - bb->num;
     }
 }
 int* twoSum(int* nums, int numsSize, int target) {

     struct node* a = (struct node*)malloc(numsSize*(sizeof(struct node)));
     int * vis = (int *)malloc(*sizeof(int));
     int beg = ,end = numsSize-,i;
     for(i = ; i < numsSize; i++) {
         a[i].num = nums[i];
         a[i].local = i+;
     }
     qsort(a,numsSize,sizeof(a[]),cmp);
     for(; beg < end;){
         int ans = target - a[beg].num;
         while(){
             if(a[end].num == ans){
                 if(a[beg].local < a[end].local){
                     vis[] = a[beg].local;
                     vis[] = a[end].local;
                 }else {
                     vis[] = a[end].local;
                     vis[] = a[beg].local;
                 }
                 return vis;
             }else if(a[end].num < ans){
                 beg++;
                 break;
             }else {
                 end--;
             }
         }
     }
     free(a);
 }

(2)用c++提供的map模拟hash：

 class Solution {
 public:
     vector<int> twoSum(vector<int>& nums, int target) {
         map<int,int>a;
         int i,len;
         len = nums.size();
         for(i = ; i < len; i++) {
             int ans = target - nums[i];
             if(a.count(nums[i])){
                 vector<int> res;
                 int ans1,ans2;
                 ans1 = i+;
                 ans2 = a[nums[i]];
                 if(ans1 > ans2) {
                     res.push_back(ans2);
                     res.push_back(ans1);
                     return res;
                 }else {
                     res.push_back(ans1);
                     res.push_back(ans2);
                     return res;
                 }
             }
             a[ans] = i+;
         }
     }
 };