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[Leetcode]Two Sum

时间:2015-06-24 10:59:38      阅读:112      评论:0      收藏:0      [点我收藏+]

标签:leetcode   hash map   

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

开始写了一个O(n^2)的解法,超时了。
在网上看到了一种O(n)的解决方式,利用hash map。只遍历一遍数组,把遍历过的数存在hash map中,数值作为key,序号i作为value,这样的话可以实现对已遍历数值的快速查找。
例如,当前遍历到第k个元素,然后去hash map中寻找target-nums[k]这个key是否存在,如果存在打印两个坐标,如果不存在把当前key-value对存入hash map,当然保存的前提是这个值是第一次出现,如果之前已经出现过的话,就不保存直接跳到下一个元素。
在C++中当然就用map实现了,不过C++中的map默认使用堆来实现的,平均查找复杂度为O(logn)。最后时间复杂度为O(nlogn)


class Solution {
public:
     vector<int> twoSum(vector<int>& nums, int target) {
          std::vector<int> result;
          if (nums.size() == 0) return result;
          int index1 = 0, index2 = 0;
          map<int, int> appeared;
          for (int i = 0; i < nums.size(); i++){
               if (appeared.count(target - nums[i])){
                    result.push_back(appeared[target - nums[i]] + 1);
                    result.push_back(i + 1);
                    return result;
               }
               if (!appeared.count(nums[i]))
                    appeared[nums[i]] = i;
          }
          
     }
};



[Leetcode]Two Sum

标签:leetcode   hash map   

原文地址:http://blog.csdn.net/cr_peace/article/details/46618433

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