Given an array of integers nums and an integer target, return indices of the two numbers such that they add up to target.
You may assume that each input would have exactly one solution, and you may not use the same element twice.
You can return the answer in any order.
Example 1:
Input: nums = [2,7,11,15], target = 9
Output: [0,1]
Output: Because nums[0] + nums[1] == 9, we return [0, 1].
Example 2:
Input: nums = [3,2,4], target = 6
Output: [1,2]
Example 3:
Input: nums = [3,3], target = 6
Output: [0,1]
Constraints:
- 2 <= nums.length <= 104
- -109 <= nums[i] <= 109
- -109 <= target <= 109
- Only one valid answer exists.
Follow-up: Can you come up with an algorithm that is less than O(n2) time complexity?
class Solution {
public int[] twoSum(int[] nums, int target) {
Map<Integer, Integer> map = new HashMap<>();
for(int i=0;i<nums.length;i++){
if(map.containsKey(target-nums[i])){
return new int[]{map.get(target-nums[i]), i};
}else{
map.put(nums[i], i);
}
}
throw new IllegalArgumentException("no solution!");
}
}- 先排序再返回，复杂度至少为O(nlong)
- 注意返回的是数组下标，这里用map存储每个数字和下标元素