LeetCode——Permutations
2014-08-10 09:57
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Given a collection of numbers, return all possible permutations.
For example,
and
原题链接:https://oj.leetcode.com/problems/permutations/
题目:给定一组数,返回其全排列。
思路:记录每一个数是否被使用过,将未使用过的数加入到current中,current长度已满,则加入到result中。
public List<List<Integer>> permute(int[] num) {
if (num == null)
return null;
List<List<Integer>> result = new ArrayList<List<Integer>>();
if (num.length == 0)
return result;
permute(num, new boolean[num.length], new ArrayList<Integer>(), result);
return result;
}
public void permute(int[] num, boolean[] isused,
ArrayList<Integer> current, List<List<Integer>> result) {
if (current.size() == num.length) {
result.add(new ArrayList<Integer>(current));
return;
}
for (int i = 0; i < num.length; i++) {
if (!isused[i]) {
isused[i] = true;
current.add(num[i]);
permute(num, isused, current, result);
current.remove(current.size() - 1);
isused[i] = false;
}
}
}
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].
原题链接:https://oj.leetcode.com/problems/permutations/
题目:给定一组数,返回其全排列。
思路:记录每一个数是否被使用过,将未使用过的数加入到current中,current长度已满,则加入到result中。
public List<List<Integer>> permute(int[] num) {
if (num == null)
return null;
List<List<Integer>> result = new ArrayList<List<Integer>>();
if (num.length == 0)
return result;
permute(num, new boolean[num.length], new ArrayList<Integer>(), result);
return result;
}
public void permute(int[] num, boolean[] isused,
ArrayList<Integer> current, List<List<Integer>> result) {
if (current.size() == num.length) {
result.add(new ArrayList<Integer>(current));
return;
}
for (int i = 0; i < num.length; i++) {
if (!isused[i]) {
isused[i] = true;
current.add(num[i]);
permute(num, isused, current, result);
current.remove(current.size() - 1);
isused[i] = false;
}
}
}
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