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C++:
vector<int> twoSum(vector<int>& nums, int target) { unordered_map<int, int> hashMap;? for (int i = 0; i < nums.size(); i++) { if(hashMap.find(nums[i]) == hashMap.end()){ hashMap[target-nums[i]] = i; }else{ return vector<int> {hashMap[nums[i]]+1, i+1}; } } return vector<int> {}; }
1. hashMap[value] = i 使得value + nums[i] = target
2. unordered_map其内部存储为hash、遍历无序、使用需重载operator ==以及hash_value()， map存储为树、需重载operator <; 详见文章http://blog.csdn.net/orzlzro/article/details/7099231
3. hashMap[nums[i]]一定比i小，因前者值为几个迭代之前的i而这里i从小到大
Python：
def twoSum(self, nums, target): m_map = {} for i in range(len(nums)): if nums[i] not in m_map: m_map[target - nums[i]] = i else: return[m_map[nums[i]]+1, i+1]
讨论里有更简洁代码，
for j, item in enumerate(nums, 1): #start from 1 & items are entries i = m_map.get(item, -1) #the same as m_map[] but instead of crush, gives back -1 when couldn‘t find item if i > 0: return [i, j] m_map[target - item] = j
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原文地址：http://www.cnblogs.com/rangozhang/p/4545222.html
