【LeetCode】1. Two Sum(两数之和)-unordered_map实现
本题是一下公司的面试题:
问题描述:
问题求解:
使用无序容器unordered_map实现:
#include <iostream>
#include <vector>
#include <cassert>
#include <unordered_map>
using namespace std;
// 时间复杂度:O(n)
// 空间复杂度:O(n)
class Solution {
public:
vector<int> twoSum(vector<int>& nums, int target) {
unordered_map<int, int> record; //record表示元素的下标
for (int i = 0; i < nums.size(); i++) {
int complement = target - nums[i]; //complement为补数
if (record.find(complement) != record.end()) { //在record中查找元素complement
int res[] = { i, record[complement] }; //将找到的两个元素的下标放到数组res中
return vector<int>(res, res + 2); //返回vector
}
record[nums[i]] = i; //将元素nums[i]放到record中,并将下标i计为record的value
}
throw invalid_argument("the input has no solution");//找不到答案,抛出异常
}
};
int main() {
const int nums[] = { 2, 7, 11, 15};
vector<int> nums_vec(nums, nums + sizeof(nums) / sizeof(int));
int target = 9;
vector<int> res = Solution().twoSum(nums_vec, target);
cout << res[0] << " " << res[1] << endl;
getchar();
return 0;
}
图解:
①当 i = 0时:
②当 i = 1 时:
Reference: