Permutation I

【题目】Given a collection of numbers, return all possible permutations.
For example,

[1,2,3]

have the following permutations:

[1,2,3]

,

[1,3,2]

,

[2,1,3]

,

[2,3,1]

,

[3,1,2]

,
and

[3,2,1]

【题意】

全排列

【分析】

遍历字符串，然后将当前遍历的字符和首字符交换位置，调用本函数处理首字符之后的子字符串，函数执行完后，再调换回来。在不需要复制字符串的情况下，避免了每次递归函数相互之间的影响。实现：

public ArrayList<ArrayList<Integer>> permutations(int[] num) {
ArrayList<ArrayList<Integer>> result = new ArrayList<ArrayList<Integer>>();
permute(num, 0, result);
return result;
}void permute(int[] num, int start, ArrayList<ArrayList<Integer>> result) {if (start >= num.length) {
ArrayList<Integer> item = convertArrayToList(num);
result.add(item);
}for (int j = start; j <= num.length - 1; j++) {
swap(num, start, j);
permute(num, start + 1, result);
swap(num, start, j);
}
}private ArrayList<Integer> convertArrayToList(int[] num) {
ArrayList<Integer> item = new ArrayList<Integer>();
for (int h = 0; h < num.length; h++) {
item.add(num[h]);
}
return item;
}private void swap(int[] a, int i, int j) {
int temp = a[i];
a[i] = a[j];
a[j] = temp;
}