Leetcode 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, 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].

題解：
1.可以用暴力的方法來解決，程式碼如下：

class Solution {
	public int[] twoSum(int[] nums, int target) {
   	 	int len = nums.length;
    	int[] a = new int[2];
   	 	for(int i = 0 ; i < len; i++){
        	for(int j = i + 1; j < len ; j++){
            	if(nums[i] + nums[j] == target){
                	a[0] = i;
                	a[1] = j;
                	break;
            	}
        	}
    	}
    	return a;
	}
}

時間複雜度：O（n2）
空間複雜度：O（1）
2.使用HashMap來做

	class Solution {
	public int[] twoSum(int[] nums, int target) {
    	int len = nums.length;
    	Map<Integer,Integer> map = new HashMap<Integer,Integer>();
    	for(int j = 0 ; j < len;j++){
        	int complement = target - nums[j];
        	if(map.containsKey(complement) && map.get(complement) < len){
            	return new int[]{map.get(complement),j};
        	}
        	map.put(nums[j],j);
    	}
    	throw new IllegalArgumentException("No two sum solution");
    }
}

時間複雜度：O（n）
空間複雜度：O（n）