1. Two Sum

Given an array of integers, return indices of the two numbers such that they add up to a specific target.

从给定序列中找出和为指定数字的两个数字的下标

You may assume that each input would have exactly one solution.

假设每个输入只有一个解

Example:

Given nums = [2, 7, 11, 15], target = 9,

Because nums[0] + nums[1] = 2 + 7 = 9,
return [0, 1].

Brute Force

time complexity O(n²)

space complexity O(1)

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

Hash

time complexity O(n)

space complexity O(n)

该方法给出的序列是从后往前的

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

see discussion here https://discuss.leetcode.com/topic/3294/accepted-c-o-n-solution/2

来自为知笔记(Wiz)