### leetcode Question 61: Next permutation

Next permutation

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.
1,2,3 → 1,3,2
3,2,1 → 1,2,3
1,1,5 → 1,5,1

Analysis:
Well, this is more like a math problem, and I don't know how to solve it.
From the wikipedia, one classic algorithm to generate next permutation is:
Step 1: Find the largest index k, such that A[k]<A[k+1]. If not exist, this is the last permutation. (in this    problem just sort the vector and return.)
Step 2: Find the largest index l, such that A[l]>A[k].
Step 3: Swap A[k] and A[l].
Step 4: Reverse A[k+1] to the end.

OK! Having the algorithm then coding becomes an easy thing!

Code(C++):
class Solution {
public:
void nextPermutation(vector<int> &num) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
int sz = num.size();
int k=-1;
int l;
//step1
for (int i=0;i<sz-1;i++){
if (num[i]<num[i+1]){
k=i;
}
}
if (k==-1){
sort(num.begin(),num.end());
return;
}

//step2
for (int i=k+1;i<sz;i++){
if (num[i]>num[k]){
l=i;
}
}
//step3
int tmp = num[l];
num[l]=num[k];
num[k]=tmp;
//step4
reverse(num.begin()+k+1,num.end());
}
};


Code(Python):

class Solution:
# @param num, a list of integer
# @return a list of integer
def nextPermutation(self, num):
k = -1;
#Step 1: find max k A[k]<A[k+1]
for i in xrange(0,len(num)-1):
if num[i]<num[i+1]:
k = i;
if k == -1:
num.reverse();
return num;
else:
#Step 2: find l A[l]>A[k]
for i in xrange(k+1, len(num)):
if num[i]>num[k]:
l = i;
#Step 3: swap A[l] A[k]
num[l], num[k] = num[k], num[l];
#Step 4: reverse k+1 to end
num[k+1:len(num):1] = num[len(num)-1:k:-1];
return num;



1. line 16: actually no need to sort, just reverse the whole array. O(lg(n)) to O(n).

2. line 21: int i could start from k+1 instead of 0

1. Thanks for your suggestion !

3. Nice! As we need the largest indices in step 1 & 2, we can loop backward for A[k - 1] < A[k] and A[l] > A[k] and break it earlier. :)

4. O(n) Java Code

private static void reverseArray(int[] num, int index) {
for (int i = 0; i < (num.length - index) / 2; i++) {
int temp = num[i + index];
num[i + index] = num[num.length - i - 1];
num[num.length - i - 1] = temp;
}
}

public static void nextPermutation(int[] num) {
int pivot = num.length - 2;;
for (; pivot >= 0; pivot --) {
if (num[pivot] < num[pivot + 1]) break;
}
System.out.println("Pivot = " + pivot);
if (pivot == -1) {
reverseArray(num, 0);
} else {
int min_index = -1;
for (int i = pivot + 1; i < num.length; i++) {
if (num[pivot] < num[i] && (min_index == -1 || num[i] <= num[min_index])) min_index = i;
}
int temp = num[pivot];
num[pivot] = num[min_index];
num[min_index] = temp;
reverseArray(num, pivot + 1);
}
}

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