classSolution {publicvoidnextPermutation(int[] nums) {//高位为nums[0]if(nums !=null&& nums.length >1){inti;for(i = nums.length-2;i>=0;i--){if(nums[i+1]>nums[i]){break; } }if(i >= 0){//如果整个序列为逆序时,i小于0 reverse整个序列,否则找到比nums[i]大的交换次序intk;for(...
publicclassSolution {publicvoidnextPermutation(int[] nums) {if(nums.length < 2)return;for(inti = nums.length - 2 ; i >= 0 ; i--) {if(nums[i] < nums[i+1]) {intindex = nums.length - 1;for(intj = i + 1 ; j < nums.length ; j++) {//找到比nums[index]大的当中最小的if...
全排列函数next_permutation(start,end),start和end是要求的序列的范围,左闭右开,这个函数是直接运行它的“下一个序列”(即字典序递增),所以最好用do–while来运行 如:用while来运行,原本1,2,3这个序列没有输出,这个代码看着有输出1,2,3,是因为前面有个printf单独输出了。 如果用do–while来... ...
The replacement must be in-place, do not allocate extra memory. Solution public class Solution { public void nextPermutation(int[] nums) { int len = nums.length; if (nums == null || len == 0) return; //从末位向前遍历,(假设循环开始全是倒序排列,如65321) for (int i = len - 2; ...
class Solution { public void nextPermutation(int[] nums) { int n = nums.length; int i = n - 2; while(i>=0 && nums[i]>=nums[i+1]){ i--; } if(i>=0){ for(int j=n-1; j>=0; j--){ //这里不能取等于 if(nums[j] > nums[i]){ ...
1publicclassSolution {2publicvoidnextPermutation(int[] num) {3//Start typing your Java solution below4//DO NOT write main() function5intj,i;6for(i = num.length - 1; i > 0; i --){7j = i - 1;8if(num[j] <num[i]){9intex = 0;10inta;11for(a = i; a < num.length; a...
Solution 3 参照LeetCode上的solution 思路 代码 31. Next Permutation 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 ascendin...
class Solution { public: void nextPermutation(vector<int> &num) { in... 42750 LeetCode 0031 - Next Permutation Next Permutation Desicription Implement next permutation, which rearranges numbers into the lexicographically...next greater permutation of numbers...1,1,5 → 1,5,1 Solution class...
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 ...
class Solution { public: void nextPermutation(vector<int> &num) { in... 42750 HarmonyOS NEXT - 通用属性 10510 Leetcode 31 Next Permutation Implement next permutation, which rearranges numbers into the lexicographically next greater permutation...偷懒做法 class Solution { public: void nextPermutatio...