【leetcode-1】Two Sum
来源:互联网 发布:微赞淘宝客浏览器失败 编辑:程序博客网 时间:2024/04/20 05:18
要求:
Given an array of integers, find two numbers such that they add up to a specific target number.
The function twoSum should return indices(index的复数,索引) 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
线索:
numbers[index1]+numbers[index2]=target
index1 < index2
代码:
解决方案一:暴力搜索
时间复杂度:O(n*n)——n*(n-1)/2
结果:Time Limit Exceeded
代码:
public class Solution { public int[] twoSum(int[] numbers, int target) { int[] index = new int[2];//int[] index=new int[]{1,2} int[] index={1,2} for (index[0] = 0; index[0] < numbers.length - 1; index[0]++) { for (index[1] = index[0] + 1; index[1] < numbers.length; index[1]++) { if (numbers[index[0]] + numbers[index[1]] == target) break; } } return index; }}
解决方案二:哈希查找
时间复杂度:O(n)——n+n
结果:Accepted
public class Solution { public int[] twoSum(int[] numbers, int target) { int[] res = new int[2]; Map map = new HashMap(); for (int i = 0; i < numbers.length; i++) { map.put(numbers[i], i); } for (int i = 0; i < numbers.length; i++) { int gap = target - numbers[i]; if (map.get(gap) != null && (Integer)map.get(gap) != i) { res[0] = i + 1; res[1] = (Integer)map.get(gap) + 1; break; } } return res; } }
解决方案三:
public int[] twoSum(int[] numbers, int target) { int[] res=new int[2]; if(numbers==null||numbers.length<2) return res; HashMap<Integer,ArrayList<Integer>> hm=new HashMap<Integer,ArrayList<Integer>>(); for(int i=0;i<numbers.length;i++) { if(hm.containsKey(numbers[i])) { hm.get(numbers[i]).add(i); } else { ArrayList<Integer> indexArr=new ArrayList<Integer>(); indexArr.add(i); hm.put(numbers[i],indexArr); } } for(int i=0;i<numbers.length;i++) { if(hm.containsKey(target-numbers[i])) { boolean isDouble=target==2*numbers[i]; if(isDouble&&hm.get(numbers[i]).size()!=1) { res[0]=hm.get(numbers[i]).get(0)+1; res[1]=hm.get(numbers[i]).get(1)+1; break; } else if((!isDouble)&&hm.get(numbers[i]).size()==1) { res[0]=hm.get(numbers[i]).get(0)+1; res[1]=hm.get(target-numbers[i]).get(0)+1; break; } } } return res;}
评语:
O(n2) runtime, O(1) space – Brute force:简单匹配算法
The brute force approach is simple. Loop through each element x and find if there is another value that equals to target – x. As finding another value requires looping through the rest of array, its runtime complexity is O(n2).
O(n) runtime, O(n) space – Hash table:哈希表
We could reduce the runtime complexity of looking up a value to O(1) using a hash map that maps a value to its index.
public class Solution { public int[] twoSum(int[] numbers, int target) { Map<Integer, Integer> map = new HashMap<Integer, Integer>(); for (int i = 0; i < numbers.length; i++) { int x = numbers[i]; if (map.containsKey(target - x)) { return new int[] { map.get(target - x) + 1, i + 1 }; } map.put(x, i); } throw new IllegalArgumentException("No two sum solution"); }}
Average Rating: 4.5 (87 votes)
Is this solution helpful? Read our book to learn more.
0 0
- LeetCode 1 - Two Sum
- leetcode 1 Two Sum
- Leetcode【1】:Two Sum
- [leetcode 1] Two Sum
- 【leetcode-1】Two Sum
- [leetcode 1] Two Sum
- [Leetcode] 1 - Two Sum
- LeetCode (1) Two Sum
- LeetCode 1:《Two Sum》
- LeetCode | #1 Two Sum
- leetcode-1 Two Sum
- Two Sum | LeetCode(1)
- [Leetcode]1Two Sum
- leetcode 1 Two Sum
- leetcode #1 Two Sum
- leetcode 1:Two Sum
- LeetCode 1 Two Sum
- leetcode #1 two sum
- Nc57二次开发合同先进先出
- CDbCriteria 详细便用说明
- sin和圆形弧度制的表示
- Java中如何封装自己的类,建立并使用自己的类库?
- UVa 628 - Passwords
- 【leetcode-1】Two Sum
- YII 快速创建项目GII
- canal安装部署
- eclipse调试 Red5 项目时查看 Red5 容器内部运行机制
- ios学习之UITableView(一)
- Javascript 面向对象编程(一):封装
- out类类型数组
- 找不到答案的时候,就独自出去看一看这个世界
- 中断线程的一个方法