1. Two Sum
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.
Example:
Given nums = [2, 7, 11, 15], target = 9, Because nums[0] + nums[1] = 2 + 7 = 9, return [0, 1].
题目大意：
在一个数组中找出2个元素的和等于目标数，输出这两个元素的下标。
思路：
最笨的办法喽，双循环来处理。时间复杂度O(n*n)。
代码如下：
|
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
|
class
Solution {
public
:
vector<
int
> twoSum(vector<
int
>& nums,
int
target) {
vector<
int
> result;
int
i,j;
for
(i = 0; i < nums.size();i++)
{
for
(j = i+1; j < nums.size();j++)
{
if
(nums[i] + nums[j] == target)
{
result.push_back(i);
result.push_back(j);
break
;
}
}
}
return
result;
}
};
|
参考他人的做法：https://discuss.leetcode.com/topic/3294/accepted-c-o-n-solution
采用map的键值，把元素做键，把元素的下标做值。
|
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
|
vector<
int
> twoSum(vector<
int
> &numbers,
int
target)
{
//Key is the number and value is its index in the vector.
unordered_map<
int
,
int
> hash;
vector<
int
> result;
for
(
int
i = 0; i < numbers.size(); i++) {
int
numberToFind = target - numbers[i];
//if numberToFind is found in map, return them
if
(hash.find(numberToFind) != hash.end()) {
result.push_back(hash[numberToFind]);
result.push_back(i);
return
result;
}
//number was not found. Put it in the map.
hash[numbers[i]] = i;
}
return
result;
}
|
本文转自313119992 51CTO博客，原文链接:http://blog.51cto.com/qiaopeng688/1836898
