#51 Previous Permutation

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题目描述：

Given a list of integers, which denote a permutation.

Find the previous permutation in ascending order.

Notice

The list may contains duplicate integers.

Have you met this question in a real interview?

Yes

Example

For [1,3,2,3], the previous permutation is [1,2,3,3]

For [1,2,3,4], the previous permutation is [4,3,2,1]

题目思路：

这题和#51差不多，有些地方改一下就好了。

Mycode（AC = 30ms）：

class Solution {public:    /**     * @param nums: An array of integers     * @return: An array of integers that's previous permuation     */    vector<int> previousPermuation(vector<int> &nums) {        // write your code here        if (nums.size() <= 1) return nums;                // find the key position        int idx = -1;        for (int i = nums.size() - 2; i >= 0; i--) {            if (nums[i] > nums[i + 1]) {                idx = i;                break;            }        }                // if is the larget sequence, return reverse        if (idx == -1) {            reverse(nums.begin(), nums.end());            return nums;        }                int second_max = nums[idx + 1], second_idx = idx + 1;        for (int i = idx + 1; i < nums.size(); i++) {            if (nums[i] < nums[idx] && nums[i] > second_max) {                second_max = nums[i];                second_idx = i;            }        }                // swap        swap(nums, idx, second_idx);                // sort the rest numbers with decending order        sort(nums.begin() + idx + 1, nums.end());        reverse(nums.begin() + idx + 1, nums.end());                return nums;    }        void swap(vector<int> &nums, int i, int j) {        int tmp = nums[i];        nums[i] = nums[j];        nums[j] = tmp;    }};

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