[leetcode 001]Two Sum

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Question:

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.

Example:

Given nums = [2, 7, 11, 15], target = 9,
Because nums[0] + nums[1] = 2 + 7 = 9,
return [0, 1].

Method:

Java

• 1.暴力双循环，时间复杂度为O（n^2）

1. publicclassSolution{
2. publicint[] twoSum(int[] nums,int target){
3. int[] arr =newint[2];
4. for(int i =0;i< nums.length-1; i++){
5. for(int j = i+1; j< nums.length; j++){
6. if(target == nums[i]+ nums[j]){
7. arr[0]= i ;
8. arr[1]= j ;
9. }
10. }
11. }
12. return arr ;
13. }
14. }

• 2.利用HashMap 或者 HashTable,时间复杂度为O(N)

1. publicclassSolution{
2. publicint[] twoSum(int[] nums,int target){
3. int[] result ={0,0};
4. Map<Integer,Integer> map =newHashMap<Integer,Integer>();
5. for(int i =0;i<nums.length;i++){
6. if(map.containsKey(target-nums[i])){
7. result[0]= map.get(target-nums[i]);
8. result[1]= i;
9. }else{
10. map.put(nums[i],i);
11. }
12. }
13. return result ;
14. }
15. }

思路解析：
明显优于第一种解法，时间复杂度上大大降低
创建一个hash表，将数组的v值和索引值当作hash表的K-V
当存在 target-k存在于表中时，即取出对应的value值，就是所要找的位置

Python
同样的Hashtable思路，key is number，value is index

1. classSolution(object):
2. def twoSum(self, nums, target):
3. """
4. :type nums: List[int]
5. :type target: int
6. :rtype: List[int]
7. """
8. index ={};
9. for i,j in enumerate(nums):
10. if target-j in index:
11. return[index[target-j],i]
12. index[j]= i

[leetcode 001]Two Sum

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原文地址：http://www.cnblogs.com/zhjiann/p/5495491.html