2.1.8 3Sum

原創

2020-02-21 23:54

Link: https://oj.leetcode.com/problems/3sum/

這題和2sum差不多，但“使用哈希表的解法並不是很方便，因爲結果數組中元素可能重複，如果不排序對於重複的處理將會比較麻煩”//? Ref:http://blog.csdn.net/linhuanmars/article/details/19711651

Approach: 先排序，再左右夾逼。

Time: O(n^2+nlogn)=(n^2), Space: O(1)

但要注意：題目中說不允許出現duplicate triplets。有兩種方法處理：

1 先用hashset存結果，再轉到arraylist裏

public class Solution {
    public ArrayList<ArrayList<Integer>> threeSum(int[] num) {
        ArrayList<ArrayList<Integer>> result = new ArrayList<ArrayList<Integer>>();
        HashSet<ArrayList<Integer>> set = new HashSet<ArrayList<Integer>>();
        Arrays.sort(num);
        for(int i = 0; i < num.length; i++){
            int j = i+1; 
            int k = num.length - 1;
            while(j < k){
                int sum = num[j] + num[k];
                if(sum == -num[i]){
                    ArrayList<Integer> list = new ArrayList<Integer>();
                    list.add(num[i]);
                    list.add(num[j]);
                    list.add(num[k]);
                    set.add(list);
                    j++;//there may be multiple triplets, so do not return here
                    k--;//there may be multiple triplets, so do not return here
                }
                else if ((sum < -num[i])){
                    j++;
                }
                else {
                    k--;
                }
            }
        }
        for(ArrayList<Integer> item : set){
            result.add(item);
        }
        return result;
    }
}

2 跳過值相等的元素 (見diff1)。比如數組裏有兩個3，對第一個3遍歷過以後，第二個3就不用處理了，因爲所有和等於-3的都已經處理過了。

public class Solution {
    public ArrayList<ArrayList<Integer>> threeSum(int[] num) {
        ArrayList<ArrayList<Integer>> result = new ArrayList<ArrayList<Integer>>();
        Arrays.sort(num);
        for(int i = 0; i < num.length; i++){
            if(i!=0 && num[i] == num[i-1]) continue;//diff1
            int j = i+1; 
            int k = num.length - 1;
            while(j < k){
                int sum = num[j] + num[k];
                if(sum == -num[i]){
                    ArrayList<Integer> list = new ArrayList<Integer>();
                    list.add(num[i]);
                    list.add(num[j]);
                    list.add(num[k]);
                    result.add(list);//diff2
                    j++;
                    k--;
                   while(j < k && num[j] == num[j-1]){//diff3
                       j++;
                    }
                    while(j < k && num[k] == num[k+1]){//diff4
                        k--;
                    }
                }
                else if ((sum < -num[i])){
                    j++;
                }
                else {
                    k--;
                }
            }
        }
        return result;
    }
}

Note: Both Approaches use this for loop to avoid duplicate triplets

 for(int i = 0; i < num.length; i++){
            int j = i+1; 
            int k = num.length - 1;

2015.1.31

Must have the the if and two whiles to avoid duplicates:

if(i!=0 && num[i] == num[i-1]) continue;//diff1

                 while(j < k && num[j] == num[j-1]){//diff3
                    j++;
                 }
                 while(j < k && num[k] == num[k+1]){//diff4
                        k--;
                 }