# [2165. Smallest Value of the Rearranged Number](https://leetcode.com/problems/smallest-value-of-the-rearranged-number/)


## 题目

You are given an integer `num.` **Rearrange** the digits of `num` such that its value is **minimized** and it does not contain **any** leading zeros.

Return *the rearranged number with minimal value*.

Note that the sign of the number does not change after rearranging the digits.

**Example 1:**

```
Input: num = 310
Output: 103
Explanation: The possible arrangements for the digits of 310 are 013, 031, 103, 130, 301, 310.
The arrangement with the smallest value that does not contain any leading zeros is 103.

```

**Example 2:**

```
Input: num = -7605
Output: -7650
Explanation: Some possible arrangements for the digits of -7605 are -7650, -6705, -5076, -0567.
The arrangement with the smallest value that does not contain any leading zeros is -7650.

```

**Constraints:**

- `10^15 <= num <= 10^15`

## 题目大意

给你一个整数 num 。重排 num 中的各位数字，使其值 最小化 且不含 任何 前导零。

返回不含前导零且值最小的重排数字。注意，重排各位数字后，num 的符号不会改变。

## 解题思路

- 先将每个数字出现次数统计出来。然后将数字大小从小到大排序。如果原数是正数，当出现有数字 0 的情况的时候，需先将第二小的数字排列到第一个，再把 0 排列完。再继续排列第二小，第三小。。。
- 如果原数是负数。那么就逆序排列，即先排列最大的数字，然后次大的数字，直到排列最小的数字。因为数字越大，对应的这个数的负数就越小。

## 代码

```go
package leetcode

import "sort"

func smallestNumber(num int64) int64 {
	pos := true
	if num < 0 {
		pos = false
		num *= -1
	}
	nums, m, res := []int{}, map[int]int{}, 0
	for num != 0 {
		tmp := int(num % 10)
		m[tmp]++
		num = num / 10
	}

	for k := range m {
		nums = append(nums, k)
	}
	if pos {
		sort.Ints(nums)
	} else {
		sort.Sort(sort.Reverse(sort.IntSlice(nums)))
	}

	if nums[0] == 0 && len(nums) > 1 {
		res += nums[1]
		m[nums[1]]--
	}

	for _, v := range nums {
		if res != 0 {
			for j := m[v]; j > 0; j-- {
				res = res * 10
				res += v
			}
		} else {
			res += v
			tmp := m[v] - 1
			for j := tmp; j > 0; j-- {
				res = res * 10
				res += v
			}
		}
	}
	if !pos {
		return -1 * int64(res)
	}
	return int64(res)
}
```