leetcode_1twosum
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题目:
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].题目没什么,主要是一句话:不能用同一个元素两次。换句话说不能把原数组循环两次。但它又没说是O(n),那可以说,我存在其他地方这个元素也只用了一次呀。。但这个题的本身目的肯定是让写O(n)的代码的。也很简单,用map。过一个元素,存进map一个值。然后过下一个元素看看map里有没有想要的值。有就返回这俩的下表。贴代码。
class Solution {public: vector<int> twoSum(vector<int>& nums, int target) { vector<int> result; map<int, int> hash; int flag = 0; for(int i = 0; i < nums.size(); i++){ if(hash.find(target - nums[i]) != hash.end()){ result.push_back(hash[target - nums[i]]); result.push_back(i); break; } hash[nums[i]] = i; } return result; }};嗯~
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