Java-Largest Number

标签：快速排序   string   

Given a list of non negative integers, arrange them such that they form the largest number.

For example, given [3, 30, 34, 5, 9], the largest formed number is 9534330.

Note: The result may be very large, so you need to return a string instead of an integer.

Credits:
Special thanks to @ts for adding this problem and creating all test cases.

给一个乱序数组 要求将其依次排列其能组成的最大数

基本要求还是排序 只是比较大小的依据不同 比如412<41但415>41比较的依据是谁放在后面组成的数更大 所以 只要将n1和n2都转换为String 分别前后放置  比较这两个组成String的大小 返回结果 排序方法采用的是快速排序 一开始在sort()函数部分 进行递归时 i错写为mid导致 死循环 花费了好长时间才发现 代码如下：

public class Solution {
    public static String largestNumber(int[] num) {		
		String ss="";
		if(num.length==0){
			return ss;
		}
		sort(num, 0, num.length-1);
		if(num[num.length-1]==0)return "0";
		for(int i=num.length-1;i>=0;i--){
			ss+=String.valueOf(num[i]);
		}
		return ss;
	}
	public static void sort(int[] num,int left,int right){
		if(right-left<10){
			for(int i=left;i<right;i++){
				for(int j=i;j<=right;j++){
					if(bigger(num[i], num[j])==1){
						int tmp=num[j];
						num[j]=num[i];
						num[i]=tmp;
					}
				}
			}
		}else{
			int mid=mid(num, left, right);
			int i=left;
			int j=right-1;
			for(;;){
				while(bigger(num[++i],mid)==-1){};
				while(bigger(num[--j],mid)==1){};
				if(i<j){
					swap(num, i, j);
				}else{
					break;
				}
			}
			swap(num, i, right-1);
			sort(num, left, i-1);
			sort(num, i+1,right);
		}
	}
	public static int mid(int[] num,int left,int right){
		int mid=(left+right)/2;
		if(bigger(num[left],num[mid])==1){
			swap(num, left, mid);
		}
		if(bigger(num[mid],num[right])==1){
			swap(num, right, mid);
		}
		if(bigger(num[left],num[mid])==1){
			swap(num, left, mid);
		}
		swap(num, mid, right-1);
		return num[right-1];
	}
	public static void swap(int[] num,int i,int j){
		int tmp=num[j];
		num[j]=num[i];
		num[i]=tmp;
	}
	public static int bigger(int num1,int num2){
		String ss1=String.valueOf(num1)+String.valueOf(num2);;
		String ss2=String.valueOf(num2)+String.valueOf(num1);
		for(int i=0;i<ss1.length();i++){
			if(ss1.charAt(i)>ss2.charAt(i)){
				return 1;
			}else{
				if(ss1.charAt(i)<ss2.charAt(i)){
					return -1;
				}
			}
		}
		return 0;
	}
}

Java-Largest Number

标签：快速排序   string   

原文地址：http://blog.csdn.net/u012734829/article/details/42740561