JavaArithmetic/src/LeetCode/LeetCode77.java at master · wpq2014/JavaArithmetic · GitHub

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package LeetCode;

import java.util.ArrayList;

import java.util.LinkedList;

import java.util.List;

public class LeetCode77 {

//https://leetcode-cn.com/problems/combinations/description/

/// 时间复杂度: O(n^k)

/// 空间复杂度: O(k)

private ArrayList<List<Integer>> res;

public List<List<Integer>> combine(int n, int k) {

res = new ArrayList<>();

if(n <= 0 || k <= 0 || k > n)

return res;

LinkedList<Integer> c = new LinkedList<>();

generateCombinations(n, k, 1, c);

return res;

}

// 求解C(n,k), 当前已经找到的组合存储在c中, 需要从start开始搜索新的元素

private void generateCombinations(int n, int k, int start, LinkedList<Integer> c){

if(c.size() == k){

res.add((List<Integer>)c.clone());

return;

}

for(int i = start ; i <= n ; i ++){

c.addLast(i);

generateCombinations(n, k, i + 1, c);

c.removeLast();

}

return;

}

// 求解C(n,k), 当前已经找到的组合存储在c中, 需要从start开始搜索新的元素

private void generateCombinations2(int n, int k, int start, LinkedList<Integer> c){

if(c.size() == k){

res.add((List<Integer>)c.clone());

return;

}

// 还有k - c.size()个空位, 所以, [i...n] 中至少要有 k - c.size() 个元素

// i最多为 n - (k - c.size()) + 1

// 这种优化叫做截枝

for(int i = start ; i <= n - (k - c.size()) + 1 ; i ++){

c.addLast(i);

generateCombinations2(n, k, i + 1, c);

c.removeLast();

}

return;

}

private static void printList(List<Integer> list){

for(Integer e: list)

System.out.print(e + " ");

System.out.println();

}

public static void main(String[] args) {

List<List<Integer>> res = (new LeetCode77()).combine(4, 2);

for(List<Integer> list: res)

printList(list);

}

JavaArithmetic/src/LeetCode/LeetCode77.java at master · wpq2014/JavaArithmetic · GitHub

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