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].My code（哈希函数）
class Solution {
public: vector<int> twoSum(vector<int>& nums, int target) { vector<int> res(2, -1); unordered_map<int, int> hashmap; // key is the number, val is index for (int i = 0; i < nums.size(); ++i) { auto got = hashmap.find(target - nums[i]); if (got != hashmap.end()) { res[1] = i; res[0] = got->second; break; } hashmap[nums[i]] = i; } return res; } };
Attention:
1.采用一种哈希函数解决冲突的做法：在产生冲突的地方增加一个单链表。
2.笔者自己编写了哈希长度为10，根据余数求解的方式，没有C++自带的hash库的效率高。
Other solution:
<span style="font-family:SimSun;">int* twoSum3(int* nums,int numsSize,int target) { int* arr; int i=0,j=1; arr = (int *)malloc(sizeof(int)*100); for(;j<=numsSize-1;j++) { <span style="white-space:pre"> </span>if(nums[i]+nums[j]==target) { arr[0]=i; arr[1]=j; break; } <span style="white-space:pre"> </span>if(j==numsSize-1) { i++; j=i; } }<span style="font-family: Menlo, Monaco, Consolas, 'Courier New', monospace; line-height: 1.42857143;">在讨论区看到的该代码，该code的作者说该代码beat 80%。</span>
