7.3 排列

46. 全排列

注意我们这次每次都重复去取元素，所以不需要starIndex，但是需要记录每个元素是否被用过，所以加入一个boolean[] used

class Solution {
    List<List<Integer>> result = new ArrayList<>();
    List<Integer> path = new ArrayList<>();
    private void backTracking(int[] nums,boolean[] used){
        if (path.size() == nums.length){
            result.add(new ArrayList<>(path));
            return;
        }
        for (int i = 0; i < nums.length; i++) {
             if (used[i] == true){//true说明已经被用过，跳过
                 continue;
             }
            used[i] = true;//因为用了，标志true
            path.add(nums[i]);
            backTracking(nums,used);
            used[i] = false;//回溯，重新标志false
            path.remove(path.size()-1);
        }
    }
    public List<List<Integer>> permute(int[] nums) {
        boolean[] used = new boolean[nums.length];
        backTracking(nums,used);
        return result;
    }
}

// 解法2：通过判断path中是否存在数字，排除已经选择的数字
class Solution {
    List<List<Integer>> result = new ArrayList<>();
    LinkedList<Integer> path = new LinkedList<>();
    public List<List<Integer>> permute(int[] nums) {
        if (nums.length == 0) return result;
        backtrack(nums, path);
        return result;
    }
    public void backtrack(int[] nums, LinkedList<Integer> path) {
        if (path.size() == nums.length) {
            result.add(new ArrayList<>(path));
        }
        for (int i =0; i < nums.length; i++) {
            // 如果path中已有，则跳过
            if (path.contains(nums[i])) {
                continue;
            } 
            path.add(nums[i]);
            backtrack(nums, path);
            path.removeLast();
        }
    }
}

47. 全排列 II

class Solution {
    List<List<Integer>> result = new ArrayList<>();
    List<Integer> path = new ArrayList<>();
    private void backTracking(int[] nums, boolean[] used) {
        if (nums.length == path.size()) {
            result.add(new ArrayList<>(path));
            return;
        }
        for (int i = 0; i < nums.length; i++) {
            if (i > 0 && nums[i] == nums[i - 1] && used[i - 1] == false) continue;
            if (used[i] == false) {
                used[i] = true;
                path.add(nums[i]);
                backTracking(nums, used);
                path.remove(path.size() - 1);
                used[i] = false;
            }
        }
    }
    public List<List<Integer>> permuteUnique(int[] nums) {
        Arrays.sort(nums);
        boolean[] used = new boolean[nums.length];
        backTracking(nums, used);
        return result;
    }
}