LeetCode:1 Two Sum(Week 1)

1. Two Sum

• 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].

• 解题思路

  • 暴力破解，遍历每个元素x并查找是否有另一个值等于target - x。
  • 时间复杂度为O(n^2)

• 代码

  class Solution {
  public:
      vector<int> twoSum(vector<int>& nums, int target) {
          vector<int> ans;
          for(int i = 0; i < nums.size() - 1; ++i) {
              for(int j = i + 1; j < nums.size(); ++j) {
                  if(target - nums[i] == nums[j]) {
                      ans.push_back(i);
                      ans.push_back(j);
                      return ans;
                  }
              }
          }
      }
  };

• 更优解法

  • 这个是看了答案发现的，利用map的查找，节省时间。
  • 时间复杂度为O(n*logn)

  class Solution {
  public:
      vector<int> twoSum(vector<int>& nums, int target) {
          vector<int> ans;
          map<int, int> myMap;
          for(int i = 0; i < nums.size(); ++i) {
              if(myMap.count(target - nums[i])) {
                  ans.push_back(myMap.find(target - nums[i])->second);
                  ans.push_back(i);
                  return ans;
              } 
              else
                  myMap.insert(make_pair(nums[i], i));
          }
      }
  };