LeetCode


题目:

Implement next permutation, which rearranges numbers into the lexicographically next greater permutation of numbers.

If such arrangement is not possible, it must rearrange it as the lowest possible order (ie, sorted in ascending order).

The replacement must be in-place, do not allocate extra memory.

Here are some examples. Inputs are in the left-hand column and its corresponding outputs are in the right-hand column.
1,2,31,3,2
3,2,11,2,3
1,1,51,5,1

思路:

摘自Stack Overflow:

Find the highest index i such that s{i} < s{i+1}. If no such index exists, the permutation is the last permutation. Find the highest index j > i such that s{j} > s{i}. Such a j must exist, since i+1 is such an index. Swap s{i} with s{j}. Reverse all the order of all of the elements* after index i.

package string;

public class NextPermutation {

public void nextPermutation(int[] nums) {
int len;
if (nums == null || (len = nums.length) < 2) return;
int index1 = -1;
for (int i = 0; i < len - 1; ++i) {
if (nums[i + 1] > nums[i])
index1
= i;
}

if (index1 != -1) {
int index2 = index1 + 1;
for (int j = index2 + 1; j < len; ++j) {
if (nums[j] > nums[index1])
index2
= j;
}

swap(nums, index1, index2);
reverse(nums, index1
+ 1, len);
}
else {
reverse(nums,
0, len);
}
}

private void swap(int[] nums, int i, int j) {
int tmp = nums[i];
nums[i]
= nums[j];
nums[j]
= tmp;
}

private void reverse(int[] nums, int start, int end) {
for (int i = start; i < (end - start) / 2 + start; ++i) {
swap(nums, i, end
- i + start - 1);
}
}

public static void main(String[] args) {
// TODO Auto-generated method stub
NextPermutation n = new NextPermutation();
int[] nums = { 1,3,2 };
n.nextPermutation(nums);
for (int i = 0; i < nums.length; ++i)
System.out.println(nums[i]);
}

}

 

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本文转载自:http://www.cnblogs.com/shuaiwhu/p/5066680.html     作者:Next Permutation - NULL00     发布日期:2015/12/22     本站转载的文章为个人学习借鉴使用,本站对版权不负任何法律责任。如果侵犯了您的隐私权益,请联系我们删除。


 
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