Two Sum

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, 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].

开始做没有别的方法，就是暴力解决：

public int[] twoSum(int[] nums, int target) {
int []res = new int[2];
for(int i=0;i<nums.length;i++){
int exp = target-nums[i];
for(int j = i+1;j<nums.length;j++){
if(nums[j]==exp){
res[0] = i;
res[1] = j;
break A;

}
}
}

return res;

}
}

这样的话时间复杂度O(n²),空间复杂度O(1)，

对于大数组的话运行时间比较大。

查询解决方法后，用hash表来存储，牺牲空间节省时间时间复杂度O(n),空间也是O(n)，

代码如下：

public class Solution {
public int[] twoSum(int[] nums, int target) {
int []res = new int[2];
Map<Integer,Integer> map = new HashMap<>(nums.length);
for(int i=0;i<nums.length;i++){
Integer exp = target-nums[i];

if(map.containsKey(exp)){
res[0]= map.get(exp);
res[1] = i;
break;
}
map.put(nums[i],i);
}

return res;

}
}