Next Permutation
2016-03-02 18:52
330 查看
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.
1,2,3→
1,3,2
3,2,1→
1,2,3
1,1,5→
1,5,1
方法
从后往前。找到第一个i的值比i+1值小的位置。
寻找i以后比i大的最小的数,和i交换。
将i以后的数从小到大排序。
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; } } } }
相关文章推荐
- LeetCode - 36. Valid Sudoku
- hdu2199 Can you solve this equation?(二分)
- Linux正常关机命令
- HDU 4614 Vases and Flowers (线段树 + 二分)
- ZOJ 3600 Taxi Fare(模拟)
- Tomcat的HTTP和AJP连接器
- EDA 和 SOA 的融合以及实践
- 修路方案(次小生成树)
- POJ 3667
- LeetCode 40 - Combination Sum II
- nginx的常用命令
- nyoj A*B Problem II 623 (矩阵相乘)
- Oracle 11g R2 RAC高可用连接特性 – SCAN详解2
- 【tyvj1863】【codevs1995】黑魔法师之门,有毒的并查集
- 0302感想及问题
- win7系统下安装、配置、测试nginx
- poj1149 PIGS-Dinic模板
- Linux文件系统基础(3)
- Oracle 11g R2 RAC高可用连接特性 – SCAN详解1
- PCA使用SVD解决