[LeetCode] 1053. Previous Permutation With One Swap
grandyang opened this issue · comments
Given an array of positive integers arr
(not necessarily distinct), return the lexicographically largest permutation that is smaller than arr
, that can be made with exactly one swap (A swap exchanges the positions of two numbers arr[i]
and arr[j]
). If it cannot be done, then return the same array.
Example 1:
Input: arr = [3,2,1]
Output: [3,1,2]
Explanation: Swapping 2 and 1.
Example 2:
Input: arr = [1,1,5]
Output: [1,1,5]
Explanation: This is already the smallest permutation.
Example 3:
Input: arr = [1,9,4,6,7]
Output: [1,7,4,6,9]
Explanation: Swapping 9 and 7.
Example 4:
Input: arr = [3,1,1,3]
Output: [1,3,1,3]
Explanation: Swapping 1 and 3.
Constraints:
1 <= arr.length <= 104
1 <= arr[i] <= 104
这道题给了一个正整数的数组,说是让任意交换两个数字,使得变成字母顺序最大的一种全排列,但是需要小于原先的排列,若无法得到这样的全排列(说明当前已经是最小的全排列),则返回原数组。通过分析题目中给的例子不难理解题意,根据例子2来看,若给定的数组就是升序排列的,则无法得到更小的全排列,说明只有遇到降序的位置的时候,才有可能进行交换。但是假如有多个可以下降的地方呢,比如例子1,3到2下降,2到1下降,这里是需要交换2和1的,所以最好是从后往前检验,遇到前一个数字比当前数字大的情况时,前一个数字必定是交换方之一,而当前数字并不是。比如例子3,数字4的前面是9,正确结果是9和7交换,所以还要从4往后遍历一下,找到一个仅次于9的数字交换才行,而且数字相同的话,取坐标较小的那个,比如例子4就是这种情况。其实这道题的四个例子给的真不错,基本各种情况都 cover 到了,赞一个~ 分析到这里,代码就不难写了,首先从后往前遍历,假如当前数字大于等于前一个数字,直接跳过,否则说明需要交换的。从当前位置再向后遍历一遍,找到第一个仅次于拐点的数字交换即可,注意下面的代码虽然嵌套了两个 for 循环,其实是线性的时间复杂度,参见代码如下:
class Solution {
public:
vector<int> prevPermOpt1(vector<int>& arr) {
int n = arr.size(), mx = 0, idx = -1;
for (int i = n - 1; i > 0; --i) {
if (arr[i] >= arr[i - 1]) continue;
for (int j = i; j < n; ++j) {
if (arr[j] < arr[i - 1] && mx < arr[j]) {
mx = arr[j];
idx = j;
}
}
swap(arr[i - 1], arr[idx]);
break;
}
return arr;
}
};
Github 同步地址:
参考资料:
https://leetcode.com/problems/previous-permutation-with-one-swap/
https://leetcode.com/problems/previous-permutation-with-one-swap/discuss/838836/Java-O(n)