1. Two Sum [two numbers]

description

Given an integer array nums and a target value target, and ask you to identify the target value of the two integers in the array, and return to their array subscript

Examples
given nums = [2, 7, 11 , 15], target = 9
because nums [0] + nums [1 ] = 2 + 7 = 9
is returned [0, 1]

Thinking

• Violence: two for loops
• Enhanced method (recommended, O (n) as required to return its subscript, more convenient): use the dictionary to find is O (1)

{Number needed: Who am I} through each number, if I'm not needed, I went to put my needs who wrote in the dictionary

• The third method (O (nlogn): first save the subscript of each number, the reordering O (nlogn), using two pointers (sort need to be known, and the target is greater than or less than the target to which the pointer moves)
  answers
• python

class Solution:
    def twoSum(self, nums: List[int], target: int) -> List[int]:
        # {需要的差值:index}
        d={}
        for i in range(len(nums)):
            if d.get(nums[i]) is not None: #第二个参数不写，默认返回None,注意返回0时，也会被当作false
                
                return [d.get(nums[i]), i]#i会比较大
            else:
                d[target-nums[i]]=i
        return []

• c++

class Solution {
public:
    vector<int> twoSum(vector<int>& nums, int target) {
     
        unordered_map<int,int> m;
        for (int i=0; i<nums.size(); i++)
            if (m.find(nums[i])!=m.end())
                return {m[nums[i]], i};
            else
                m[target-nums[i]]=i;
        return {};                    
    }
};