LeetCode题解2.1.7~2.1.8

2.1.7 Two Sum

描述
Given an array of integers, find two numbers such that they add up to a specific target number.

the function twoSum should return indices of the two numbers such that they add up to the target,
where index1 must be less than index2. Please note that your returned answers (both index1 and index2)
are not zero-based.
You may assume that each input would have exactly one solution.
Input: numbers={2, 7, 11, 15}, target=9

Output: index1=1, index2=2

该题使用hash表可在O(n)的时间内计算出结果

public static String solution2_1_7(int[] num,int target){HashMap<Integer,Integer> map=new HashMap<Integer,Integer>();for(int i=0;i<num.length;i++){map.put(num[i],i+1);}for(int i=0;i<num.length;i++){int temp=target-num[i];if(map.containsKey(temp)){return (i+1)+"："+map.get(temp);}}return null;}

2.1.8 3Sum

Given an array S of n integers, are there elements a; b; c in S such that a+b+c = 0? Find all unique
triplets in the array which gives the sum of zero.
Note:
• Elements in a triplet (a; b; c) must be in non-descending order. (ie, a  b  c)
• e solution set must not contain duplicate triplets.
For example, given array S = {-1 0 1 2 -1 -4}.
A solution set is:
(-1, 0, 1)
(-1, -1, 2)

对于此类题目事实上可以将0变为K，照样可以计算出结果

http://blog.csdn.net/ljiabin/article/details/40620579

<pre name="code" class="java">public static ArrayList<HashMap<String,Integer>> solution2_1_8(ArrayList<Integer> num,int target){ArrayList<HashMap<String,Integer>> result=new ArrayList<HashMap<String,Integer>>();if (num.size()<3)return null;else{Collections.sort(num);for(int i=0;i<num.size()-2;i++){if(num.get(i)==num.get(i+1))continue;find(num,i,num.size()-1,target,result);}}return result;}private static void find(ArrayList<Integer> num,int begin,int end,int target,ArrayList<HashMap<String,Integer>> result){int s=begin+1,e=end,temp=num.get(begin);while(s<e){if((temp+num.get(s)+num.get(e))==target){HashMap<String,Integer> m=new HashMap<String,Integer>();m.put("index1", begin);m.put("index2", s);m.put("index3", e);result.add(m);while(s<e&&num.get(s)==num.get(s+1))s++;while(s<e&&num.get(e)==num.get(e-1))e--;s++;e--;}else{if((temp+num.get(s)+num.get(e))>target)e--;elses++;}}}