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1.two-sum.cpp
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1.two-sum.cpp
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/*
* @lc app=leetcode id=1 lang=cpp
*
* [1] Two Sum
*
* https://leetcode.com/problems/two-sum/description/
*
* algorithms
* Easy (44.46%)
* Likes: 13943
* Dislikes: 510
* Total Accepted: 2.7M
* Total Submissions: 5.9M
* Testcase Example: '[2,7,11,15]\n9'
*
* 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, and you may
* not use the same element twice.
*
* Example:
*
*
* Given nums = [2, 7, 11, 15], target = 9,
*
* Because nums[0] + nums[1] = 2 + 7 = 9,
* return [0, 1].
*
*
*/
// @lc code=start
#include <unordered_map>
#include <vector>
using namespace std;
class Solution {
public:
// 暴力求解法,
// vector<int> twoSum(vector<int>& nums, int target) {
// vector<int> ret;
// auto size = nums.size();
// for(auto i = 0; i < size; ++i ){
// auto tmp1 = nums[i];
// for(auto j = i + 1; j < size; ++j){
// auto sum = tmp1 + nums[j];
// if(sum == target){
// ret.push_back(i);
// ret.push_back(j);
// return ret;
// }
// else{
// continue;
// }
// }
// }
// return ret;
// }
// 哈希表,先填入哈希表,然后查找
// vector<int> twoSum(vector<int>& nums, int target) {
// vector<int> ret;
// unordered_map<int,int> buf;
// for(int i = 0; i < nums.size(); ++i){
// buf[nums[i]] = i;
// }
// for(int i = 0; i < nums.size(); ++i){
// auto ele = target - nums[i];
// if(buf.count(ele) && buf[ele] != i){
// return {i,buf[ele]};
// }
// }
// return ret;
// }
// 哈希表,只填入一次
vector<int> twoSum(vector<int>& nums, int target) {
vector<int> ret;
unordered_map<int,int> buf;
for(int i = 0; i < nums.size(); ++i){
buf[nums[i]] = i;
}
for(int i = 0; i < nums.size(); ++i){
auto ele = target - nums[i];
if(buf.count(ele) && buf[ele] != i){
return {i,buf[ele]};
}
}
return ret;
}
};
// @lc code=end