Sword refers to offer 45. Arrange the array into the smallest number

Sword refers to offer 45. Arrange the array into the smallest number

Title description

Problem-solving ideas

This question seeks the smallest number to be spliced ​​together, which is essentially a sorting problem . Nums string array disposed between any two numbers x and y, the predetermined sorting rule is determined as:

• If the concatenated string x + y> y + x, then x is "greater than" y;
• Conversely, if x + y <y + x, then x is "less than" y;

x "less than" y means: after sorting, x should be to the left of y in the array; "greater than" is the opposite.

---

Added: String class compareTo()methods for comparing two strings of size, forstr1.compareTo(str2)

• If str1 == str2, the return value is 0;
• If str1 <str2, return a value less than 0;
• If str1> str2, a value greater than 0 is returned.

---

1. Arrays.sort custom sort

class Solution {
    
    
    public String minNumber(int[] nums) {
    
    
        //保存 nums 所有元素的 String 类型
        String[] stringOfNums = new String[nums.length];
        for (int i = 0; i < nums.length; i++) {
    
    
            stringOfNums[i] = String.valueOf(nums[i]);
        }

        Arrays.sort(stringOfNums, (x, y) -> (x + y).compareTo(y + x));

        StringBuilder res = new StringBuilder();
        for (String str : stringOfNums) {
    
    
            res.append(str);
        }
        return res.toString();
    }
}

2. Custom Quick Sort

class Solution {
    
    
    public String minNumber(int[] nums) {
    
    
        //保存 nums 所有元素的 String 类型
        String[] str = new String[nums.length];
        for (int i = 0; i < nums.length; i++) {
    
    
            str[i] = String.valueOf(nums[i]);
        }
        quickSort(str, 0, nums.length - 1);
        
        StringBuilder res = new StringBuilder();
        for (String sub : str) {
    
    
            res.append(sub);
        }
        return res.toString();
    }

    //自定义快速排序
    public void quickSort(String[] str, int left, int right) {
    
    
        if (left >= right) return;
        int start = left, end = right;
        String temp = str[start];
        while (start < end) {
    
    
            while (start < end && (str[end] + temp).compareTo(temp + str[end]) >= 0) end--;
            str[start] = str[end];
            while (start < end && (str[start] + temp).compareTo(temp + str[start]) <= 0) start++;
            str[end] = str[start];
        }
        str[start] = temp;
        quickSort(str, left, start - 1);
        quickSort(str, start + 1, right);
    }

}