1.md

1.Two Sum

题目链接

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

一个普通的遍历算法，时间为$O(n^2)$，速度较慢。 如果使用hashmap的方式可以使速度变快，只需要$O(n)$就可以完成 代码如下：

public class Solution {
	public int[] twoSum(int[] nums, int target) {
		int[] arr = new int[2];
		HashMap<Integer, Integer> hm = new HashMap<>();
		for(int i = 0; i < nums.length; i++){
			int another = target - nums[i];
			if(hm.containsKey(another)){
				arr[0] = hm.get(another);
				arr[1] = i;
				return arr;
			}
			hm.put(nums[i], i);
		}
		return arr;
	}   
}

python 方法，利用dict

class Solution(object):
    def majorityElement(self, nums):
		judge = 0.5
		num_dict={}
		for i in range(len(nums)):
			if nums[i] in num_dict:
				num_dict[nums[i]] = num_dict[nums[i]] + 1
			else:
				num_dict[nums[i]] = 1
		for num, counts in num_dict.items():
			if counts*1.0/len(nums) > judge: 
				return num