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Coursera Algorithms week1 Interview Questions: 3Sum in quadratic time

2024-10-31 09:22:01   209人阅读

题目要求：

Design an algorithm for the 3-SUM problem that takes time proportional to n2 in the worst case. You may assume that you can sort the n integers in time proportional to n2 or better.

分析：

《算法4》这本书提供的TwoSumFast解法为NlogN，ThreeSumFast解法为N2logN，根据课后练习，要实现3Sum复杂度为N2，建议先把2Sum复杂度实现为N。同时教材提示用排好序的数组可以实现复杂度N。我想了很久，没有发现排好序的数组对复杂度降至N有太大帮助，于是在网上搜索了下大家的做法。网上的大部分都是建议用set或map来做，我决定采用map试试，果然用map很方便。代码如下：

 1 import java.util.Arrays;
 2 import java.util.HashMap;
 3 
 4 public class TwoSumLinear {
 5     public static int count(int[] a){
 6         int cnt = 0;
 7         int n = a.length;
 8         HashMap<Integer,Integer> map = new HashMap<Integer,Integer>();
 9         for(int i =0; i<n;i++){
10             if(map.get(a[i]) == null) map.put(a[i], i);
11             Integer negIndex = map.get(-a[i]);
12             if(negIndex != null && negIndex != i){
13                 System.out.println("a["+negIndex+"]="+(-a[i])+"和a["+i+"]="+a[i]);
14                 cnt++;
15             }
16         }
17         return cnt;
18     }
19     public static void main(String[] args){
20         int[] a = { 30, -40, -20, -10, 40, 0, 10, 5 };
21         System.out.println(Arrays.toString(a));
22         System.out.println(count(a));
23     }
24 }

3Sum的作业提示可以先将数组排序，基于这个思路，结合写过的2Sum线性实现方法，写出了复杂度为N2的3Sum，个人认为实现的方式已经很精简了。

 1 import java.util.Arrays;
 2 import java.util.HashMap;
 3 
 4 public class ThreeSumQuadratic {
 5     public static int count(int[] a, int target) {
 6         Arrays.sort(a);// 数组从小到大排序，后面要使用有序数组的性质简化运算
 7         System.out.println(Arrays.toString(a));
 8         System.out.println("target="+target);
 9         int cnt = 0;
10         int n = a.length;
11         HashMap<Integer, Integer> map = new HashMap<Integer, Integer>();
12         for (int i = 0; i < n; i++) {
13             map.put(a[i], i); //以数组value为key，index为map值
14         }
15         for (int i = 0; i < n - 1; i++) {//i不会超过n-2
16             for (int j = i + 1; j < n; j++) {//j从i+1开始统计，不会超过n-1
17                 int smallValue = http://www.mamicode.com/a[i] + a[j]; //因为排好序了，所以最开始的a[i]+a[j]
18                 if (smallValue > target) //当a[i]+a[j]>target时没必要计算了，因为后续的查找就会重复
19                     break;
20                 int bigValue = http://www.mamicode.com/target-smallValue; //计算出对应的数值较大的value
21                 Integer bigIndex = map.get(bigValue); //查找数值较大的value所在的位置
22                 if (bigIndex != null && bigIndex > i && bigIndex > j) {
23                     System.out.println(
24                             "[" + i + "]=" + a[i] + ",[" + j + "]" + a[j] + ",[" + bigIndex + "]" + (bigValue));
25                     cnt++;
26                 }
27             }
28         }
29         return cnt;
30     }
31 
32     public static void main(String[] args) {
33         int[] a = { 30, -40, -20, -10, 40, 0, 10, 5 };        
34         System.out.println(count(a,0));
35     }
36 }

 

Coursera Algorithms week1 Interview Questions: 3Sum in quadratic time


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