Fair Unique PermutationsMy Submissions
16%
Accepted
Given a list of numbers with duplicate number in it. Find allunique permutations.
Example
For numbers [1,2,2] the unique permutations are:
[
[1,2,2],
[2,1,2],
[2,2,1]
]
Challenge Expand
Tags Expand <span style="color:rgba(0, 0, 0, 0);">class Solution {
/**
* @param nums: A list of integers.
* @return: A list of unique permutations.
*/
ArrayList<ArrayList<Integer>> rst = new ArrayList<ArrayList<Integer>>();
ArrayList<Integer> cur = new ArrayList<Integer>();
public ArrayList<ArrayList<Integer>> permuteUnique(ArrayList<Integer> nums) {
// write your code here
if (nums == null || nums.size() == 0) {
return rst;
}
int[] num = new int[nums.size()];
for (int i = 0; i < nums.size(); i++) {
num[i] = nums.get(i);
}
Arrays.sort(num);
permute(num, 0);
return rst;
}
void permute(int[] num, int level) {
if (level == num.length) {
rst.add(new ArrayList<Integer>(cur));
return;
}
</span><span style="color:#ff0000;">for (int i = level; i < num.length; i++) {
if (i > 0 && i > level && num[i] == num[i - 1] ||
i > level && num[i] == num[level]) {
continue;</span><span style="color:rgba(0, 0, 0, 0);">
}
swap(num, level, i);
cur.add(num[level]);
permute(num, level + 1);
cur.remove(cur.size() - 1);
swap(num, level, i);
}
}
void swap(int[] num, int i, int j) {
int temp = num[i];
num[i] = num[j];
num[j] = temp;
}
}
</span>
假設input是 [0,0,0,1,9]
i > level && num[i] == num[level] ,如果當前level指向第一個0,這句保證,不會讓0和後面的0交換
i > 0 && i > level && num[i] == num[i - 1], 如果當前level指向第二個0,那麼它所引導的所有permutation其實在level指向第一個0的時候都被遍歷完了,所以skip這倆情況,目的都是同一level不要重複取相同的數來做遍歷,只要遍歷一次就好。
最開始我我是這樣寫的:
for (int i = level; i < num.length; i++) {
if (i > 0 && num[i] == num[i - 1]) {
continue;
}
這樣當level指向第一個0的時候,它還是會和後面的0交換,會重複打印。