031 - 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

意思就是找出全排序的下一个数

比如1 2 3的全排序是

123

132

213

231

312

321

如果当前值是全排序中最大的，那么就返回最小的

int mycompar(const void *a, const void *b)
{
int *x = (int *)a;
int *y = (int *)b;
if (*x > *y) return 1;
return *x == *y? 0 : -1;
}
void nextPermutation(int* nums, int numsSize)
{
int *past = nums, i;
for (i = 1; i < numsSize; i++) {
if (nums[i] > nums[i - 1]) past = nums + i;
}
if (past == nums) {
qsort(nums, numsSize, sizeof(int), mycompar);
return;
}
int *left = past - 1;
int *right = past;
for (;past - nums < numsSize;past++)
if (*past < *right && *past > *left) right = past;
int tmp = *left;
*left = *right;
*right = tmp;
qsort(left + 1, numsSize - (left + 1 - nums), sizeof(int), mycompar);
}