Next Permutation

題目

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

方法

從後往前。找到第一個i的值比i+1值小的位置。

尋找i之後比i大的最小的數，和i交換。

將i之後的數從小到大排序。

    public void nextPermutation(int[] num) {
        if (num != null && num.length != 0 && num.length != 1) {
            int len = num.length;
            int i = len;
            for (; i > 1; i--) {
                if (num[i - 1] > num[i - 2]) {
                	
                	int flag = 0;
                	
                	for (int k = i - 1; k < len; k++) {
                		if (num[k] <= num[i - 2]) {
                			flag = k - 1;
                			break;
                		}
                	}
                	if (flag == 0) {
                		flag = len - 1;
                	}
                	int temp = num[flag];
                	num[flag] = num[i - 2];
                	num[i - 2] = temp;
                	Arrays.sort(num, i - 1, len);
                	break;
                }              
            }
            if (i == 1) {
                for (int k = 0; k < len / 2; k++) {
                    int temp = num[k];
                    num[k] = num[len - 1 - k];
                    num[len - 1 - k] = temp;
                }
            }
        }
    }