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].
此方法最直覺用雙迴圈下去一序做比較。
- 暴力、窮舉
- Run Time: 41 ms
- 時間複雜度: O(n2)
- 空間複雜度: O(1)
class Solution {
public int[] twoSum(int[] nums, int target) {
for(int i=0;i<nums.length;i++){
for(int j=i+1;j<nums.length;j++){
if(nums[i]+nums[j]==target){
int arr[]={i,j};
return arr;
}
}
}
return nums;
}
}
此種方法是利用容器 HashMap 下去實作使用 Key、Value 下去做搜尋,Key 儲存 twoSum 裡的內容,Value 儲存 twoSum 裡的索引值。
- One-pass Hash Table
- Run Time: 8 ms
- 時間複雜度 O(n)
- 空間複雜度 O(n)
class Solution {
public int[] twoSum(int[] nums, int target) {
Map<Integer,Integer> map=new HashMap<>();
int arr[]=new int [2];
for(int i=0;i<nums.length;i++) {
if(map.containsKey(target-nums[i])) {
arr[0]=map.get(target-nums[i]);
arr[1]=i;
return arr;
}else {
map.put(nums[i],i);
}
}
return arr;
}
}