(算法分析Week2)Two Sum[Easy]
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1. Two Sum[Easy]
Description
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].
给出一个整数数组和一个目标值,找到数组中和为该目标值的两个数,返回它们在数组中的位置
Solution
方法一:最容易想到的办法就是两重循环遍历数组,找到目标值即返回。复杂度为O(n²)
方法二:利用map{key : value}key不重复的特性,存储target-arr[i](即第i个位置对应的值),value即i。再利用map的find函数,如果返回值不为空,则说明找到对应的两个数,返回value即可。遍历一次复杂度O(n)
Complexity analysis
略。
Code
Solution {public: vector<int> twoSum(vector<int>& nums, int target) { vector<int> result; for (int i = 0; i < nums.size(); i++) { for (int j = i+1; j < nums.size(); j++) { if (nums[i] + nums[j] == target) { result.push_back(i); result.push_back(j); break; } } } return result; }};
Result
说点别的
如果不是要求返回对应位置,可以先排序,然后用两个“指针”,一个从前向后一个从后向前,两者之和大于target就往前,否则往后。写个伪代码:
left = 0;right = size - 1;while(left < right) {if (arr[left] + arr[right] == target))return left,right;else if ( < target) {left++;}esle if ( > target) {right--;}}
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