Next Permutation
给一个数组, 求它的下一个组合顺序上的组合. 这个是个传统算法, 先从后往前找到一个递增的index. 如果没有, 证明已经是最后一个组合了, 直接反转全部数组即可. 如果有, 从前往后找到第一个比这个index上大的数, 然后swap这两个数, 并且把这个数index, 后边的剩下所有数都反转.
public class Solution {
/**
* @param num: an array of integers
* @return: return nothing (void), do not return anything, modify num in-place instead
*/
public void reverse(int[] num, int start, int end) {
for (int i = start, j = end; i < j; i++, j--) {
int temp = num[i];
num[i] = num[j];
num[j] = temp;
}
}
public void nextPermutation(int[] num) {
// find the last increase index
int index = -1;
for (int i = num.length - 2; i >= 0; i--) {
if (num[i] < num[i + 1]) {
index = i;
break;
}
}
if (index == -1) {
reverse(num, 0, num.length - 1);
return;
}
// find the first bigger one
int biggerIndex = index + 1;
for (int i = num.length - 1; i > index; i--) {
if (num[i] > num[index]) {
biggerIndex = i;
break;
}
}
// swap them to make the permutation bigger
int temp = num[index];
num[index] = num[biggerIndex];
num[biggerIndex] = temp;
// reverse the last part
reverse(num, index + 1, num.length - 1);
}
}