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). 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 //https://segmentfault.com/a/1190000003766260 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 while(i >= 0 && nums[i] >= nums[i + 1]){ i--; } // 如果这个下降点还在数组内，我们找到一个比它稍微大一点的数替换 // 如果在之外，说明整个数组是降序的，是全局最大了 if(i >= 0){ int j = nums.length - 1; // 对于151，这种情况，要把最后面那个1跳过，所以要包含等于 while(j > i && nums[j] <= nums[i]){ j--; } swap(nums, i, j); } // 将下降点之前的部分倒序构成一个最小序列 reverse(nums, i + 1, nums.length - 1); } private void swap(int[] nums, int i, int j){ int tmp = nums[j]; nums[j] = nums[i]; nums[i] = tmp; } private void reverse(int[] nums, int left, int right){ while(left < right){ swap(nums, left, right); left++; right--; } } }