0556. Next Greater Element III (M)

Next Greater Element III (M)

题目

Given a positive integer n, find the smallest integer which has exactly the same digits existing in the integer n and is greater in value than n. If no such positive integer exists, return -1.

Note that the returned integer should fit in 32-bit integer, if there is a valid answer but it does not fit in 32-bit integer, return -1.

Example 1:

Input: n = 12
Output: 21

Example 2:

Input: n = 21
Output: -1 

Constraints:

• 1 <= n <= 2^31 - 1

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题意

给定一个数n，用n中所有位上的数字拼出另一个数m，使m恰好是大于n的最小的整数。

思路

可以看做是31. Next Permutation问题的变体，直接把n拆成单位数组成的数字，用next permutation的方法求值，注意溢出处理即可。

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代码实现

Java

class Solution {
    public int nextGreaterElement(int n) {
        int ans = 0;
        char[] digits = (n + "").toCharArray();

        int p = digits.length - 1;
        while (p >= 1 && digits[p] <= digits[p - 1]) {
            p--;
        }
        if (p == 0) return -1;

        int q = p - 1;
        p = digits.length - 1;
        while (digits[p] <= digits[q]) {
            p--;
        }

        swap(digits, p, q);
        reverse(digits, q + 1, digits.length - 1);

        for (int i = 0; i < digits.length; i++) {
            int d = digits[i] - '0';
          	// 防止溢出
            if (ans > Integer.MAX_VALUE / 10 || ans == Integer.MAX_VALUE / 10 && d > 7) {
                return -1;
            }
            ans = ans * 10 + d;
        }

        return ans;
    }

    private void swap(char[] A, int i, int j) {
        char tmp = A[i];
        A[i] = A[j];
        A[j] = tmp;
    }

    private void reverse(char[] A, int p, int q) {
        while (p < q) {
            swap(A, p++, q--);
        }
    }
}