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2014
11-18

LeetCode-3Sum[数组]

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, abc)
  • The 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)

标签: Array Two Pointers
分析

先排序,然后左右夹逼,复杂度 $O(n^2)$。

这个方法可以推广到$k$-sum,先排序,然后做$k-2$次循环,在最内层循环左右夹逼,时间复杂度是 $O(\max\{n \log n, n^{k-1}\})$。

代码1

// LeetCode, 3Sum
// 先排序,然后左右夹逼,时间复杂度O(n^2),空间复杂度O(1)
class Solution {
public:
    vector<vector<int>> threeSum(vector<int>& num) {
        vector<vector<int>> result;
        if (num.size() < 3) return result;
        sort(num.begin(), num.end());
        const int target = 0;

        auto last = num.end();
        for (auto a = num.begin(); a < prev(last, 2); ++a) {
            auto b = next(a);
            auto c = prev(last);
            while (b < c) {
                if (*a + *b + *c < target) {
                    ++b;
                } else if (*a + *b + *c > target) {
                    --c;
                } else {
                    result.push_back({ *a, *b, *c });
                    ++b;
                    --c;
                }
            }
        }
        sort(result.begin(), result.end());
        result.erase(unique(result.begin(), result.end()), result.end());
        return result;
    }
};

Java代码:

public class Solution {
    public List<List<Integer>> threeSum(int[] num) {
    	List<List<Integer>> result = new ArrayList();
    	
    	Arrays.sort(num);
    	int start,end,temp;
    	for(int i=0;i<num.length;i++){
    	    if( i!=0 && num[i]==num[i-1] )continue;     //num 1:only reserve first of all same values  
    	        int current=num[i];
    	        start=i+1;
    	        end=num.length-1;

    	    while(start<end){
    	        if(start!=i+1 && num[start]==num[start-1] ){        //num 2:only reserve first of all same values 
    	            start++;
    	            continue;
    	        }
    	        temp=num[start]+num[end];

    	        if(temp==-current){                 //find
    	        	List<Integer> list = new ArrayList<Integer>(3);
    	        	list.add(current);
    	        	list.add(num[start]);
    	        	list.add(num[end]);
    	        	result.add(list);
    	            start++;end--;
    	        }else if(temp>-current)end--;      
    	        else start++;                      
    	    }
    	}
    	return result;
    }
}

  1. L(X [0 .. M-1],Y [0 .. N-1])= 1 + L(X [0 .. M-2],Y [0 .. N-1])这个地方也也有笔误
    应改为L(X [0 .. M-1],Y [0 .. N-1])= 1 + L(X [0 .. M-2],Y [0 .. N-2])

  2. 嗯 分析得很到位,确实用模板编程能让面试官对你的印象更好。在设置辅助栈的时候可以这样:push时,比较要push的elem和辅助栈的栈顶,elem<=min.top(),则min.push(elem).否则只要push(elem)就好。在pop的时候,比较stack.top()与min.top(),if(stack.top()<=min.top()),则{stack.pop();min.pop();},否则{stack.pop();}.