LeetCode Next Permutation class Solution { public: void nextPermutation(vector<int> &num) { if (num.size() < 2) return; int i = num.size() - 2; while (i>=0 && (num[i] >= num[i+1])) i--; if (i >= 0) { int
4. 如果1步找不到这个index ,则不需要执行第二步。 View Code SOLUTION 2: https://github.com/yuzhangcmu/LeetCode_algorithm/blob/master/permutation/NextPermutation.java
然后最后想说的是,LeetCode的Discuss区里面有人用了四种解决方案解决这个问题,最让我印象深刻的是,其中一种方案只用了一条语句,因为C++的algorithm库里面有现成的next_permutation(First, last)方法解决这个问题。
23. 欢迎查看我的Github(https:///gavinfish/LeetCode-Python) 来获得相关源代码。
leetcode 31. 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)....
看一下 LeetCode 提供的动图更好理解一些。 再看这个过程,我们其实是从右向左找到第一个数字不再递增的位置,然后从右边找到一个刚好大于当前位的数字即可。 再看下代码吧。 public void nextPermutation(int[] nums) { int i = nums.length - 2; //找到第一个不再递增的位置 while (i >= 0 && nums...
http://bangbingsyb.blogspot.com/2014/11/leetcode-next-permutation.html 讲的比我清楚。 ** 总结:Array, 在草稿纸上多画画例子,就会有思路了。 ** Anyway, Good luck, Richardo! My code: publicclassSolution{publicvoidnextPermutation(int[]nums){if(nums==null||nums.length<=1)return;intchange=-1...
Next Permutation 标签: C++ 算法 LeetCode 数组 每日算法——leetcode系列 问题Next Permutation Difficulty:Medium 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 ...
[LeetCode] Next Permutation 简介:Well, in fact the problem of next permutation has been studied long ago. From the Wikipedia page, in the 14th century, a man named Narayana Pandita gi... Well, in fact the problem of next permutation has been studied long ago. From theWikipedia page, ...
public class Solution { public void nextPermutation(int[] nums) { if(nums.length <= 1){ return; } int i = nums.length - 2; // 找到第一个下降点,我们要把这个下降点的值增加一点点 // 对于511这种情况,要把前面两个1都跳过,所以要包含等于 ...