46Permutations.org

[#B] 46 Permutations - LeetCode

题目

Given a collection of distinct integers, return all possible permutations.

Example:

Input: [1,2,3]
Output:
[
  [1,2,3],
  [1,3,2],
  [2,1,3],
  [2,3,1],
  [3,1,2],
  [3,2,1]
]

思路

排列组合问题，每次从中选一个数，然后再从剩下的选一个，以此类推。我们需要记录哪些数是被用过的，每个数只能被用一次。用完之后要还原，因为下次可能顺序不一样。

Java

class Solution {
 
    List<List<Integer>> res = new ArrayList();
    boolean[] used;
    public List<List<Integer>> permute(int[] nums) {
        if (nums == null || nums.length == 0) return res;
        used = new boolean[nums.length];
        List<Integer> p = new ArrayList();
        generatePermutation(nums, 0, p);
        return res;
    }
 
    public void generatePermutation(int[] nums, int index, List<Integer> p) {
        if (index == nums.length) {
            res.add(p);
            return;
        }
     
        for (int i = 0; i < nums.length; i++) {
            if (!used[i]) {
                p.add(nums[i]);
                used[i] = true;
                // 注意这里最后一个参数，这里不应该是个引用传递，不然每次加入后都会被移除，结果就是空了
                generatePermutation(nums, index + 1, new ArrayList(p));
                p.remove(p.size() - 1);
                used[i] = false;
            }
        }
    }
}

CPP

class Solution {
public:
    vector<vector<int>> res;
    vector<bool> used;
 
    vector<vector<int>> permute(vector<int>& nums) {
        used = vector<bool>(nums.size(), false);
        vector<int> p;
        generatePermutations(nums, 0, p);
        return res;
    }
 
    void generatePermutations(vector<int>& nums, int index, vector<int> p) {
        if (index == nums.size()) {
            res.push_back(p);
            return;
        }
        for (int i = 0; i < nums.size(); i++) {
            if (!used[i]) {
                p.push_back(nums[i]);
                used[i] = true;
                generatePermutations(nums, index + 1, p);
                p.pop_back();
                used[i] = false;
            }
        }
    }
};

优化空间复杂度

class Solution {
public:
    vector<vector<int>> res;

    vector<vector<int>> permute(vector<int>& nums) {
        vector<int> track;
        backTrack(nums, 0, track);
        return res;
    }

    void backTrack(vector<int>& nums, vector<int>& track) {
        if (track.size() == nums.size()) {
            res.push_back(track);
            return;
        }
        for (int i = 0; i < nums.size(); ++i) {
            if (find(track.begin(), track.end(), nums[i]) == track.end()) {
                track.push_back(nums[i]);
                backTrack(nums, track);
                track.pop_back();
            }
        }
    }
};