难度: 简单 标题：Two Sum 求和
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].
思路
方法1 :两层循环，先选定一个数，然后遍历它之后的数，假如符合条件则返回坐标。
方法2:把traget - nums[i]放入hashmap。然后查询 nums[i] 在不在hashmap里，假如在的话就说明他和之前的某个数字加起来等于traget;
代码
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
|
public class Solution {
public int[] twoSum(int[] nums, int target) {
int a[] = new int[2];
for(int i = 0;i < nums.length -1;i++){
a[0] = i;
for(int j = i + 1;j < nums.length;j++){
a[1] = j;
if((nums[a[0]] + nums[a[1]]) == target){
return a;
}
}
}
return a;
}
}
public class Solution {
public int[] twoSum(int[] nums, int target) {
HashMap<Integer,Integer> map = new HashMap<>();
for (int i = 0; i < nums.length; i++) {
if (map.get(nums[i]) != null) {
int[] result = {map.get(nums[i]), i};
return result;
}
map.put(target - nums[i], i);
}
int[] result = {};
return result;
}
}
|
链接
https://leetcode.com/problems/two-sum/
