LeetCode 1. twoSums

C++:

vector<int> twoSum(vector<int>& nums, int target) {
    unordered_map<int, int> hashMap;

    for (int i = ; i < nums.size(); i++) {
            if(hashMap.find(nums[i]) == hashMap.end()){
                hashMap[target-nums[i]] = i;
            }else{
                return vector<int> {hashMap[nums[i]]+, i+};
            }
        }
    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