LeetCode Top Interview Questions 179. Largest Number (Java版; Medium)

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LeetCode Top Interview Questions 179. Largest Number (Java版; Medium)

题目描述

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

Example 1:

Input: [10,2]
Output: "210"
Example 2:

Input: [3,30,34,5,9]
Output: "9534330"
Note: The result may be very large, so you need to return a string instead of an integer.

第一次做; 贪心; String.compareTo(); 先将int[]转成Integer[], 也可以创建String[]; 比较器汇中不能用Integer.paeseInt(), 因为可能会溢出; 之所以使用String作为返回值, 就是因为避免过大的数会溢出的问题

/*
贪心
*/
class Solution {
    public String largestNumber(int[] nums) {
        if(nums==null || nums.length==0)
            return "";
        int n = nums.length;
        Integer[] arr = new Integer[n];
        for(int i=0; i<n; i++){
            arr[i] = nums[i];
        }
        Arrays.sort(arr, new Comparator<Integer>(){
            public int compare(Integer o1, Integer o2){
                //这里不能用Integer.paeseInt(), 因为可能会溢出
                return (""+o2+o1).compareTo(""+o1+o2);
            }
        });
        //解决[0,0]输出"00"的问题, 如果排序后第一个元素是0的话, 说明数组中全是0
        if(arr[0]==0)
            return "0";
        StringBuilder sb = new StringBuilder();
        for(Integer a : arr){
            sb.append(a);
        }
        //如何解决"00"问题
        return sb.toString();
    }
}

力扣题解,将int[]转成String[]; 同样是看排序后的数组的第一个元素是否是"0", 从而解决[0,0]输出"00"的问题

class Solution {
    private class LargerNumberComparator implements Comparator<String> {
        @Override
        public int compare(String a, String b) {
            String order1 = a + b;
            String order2 = b + a;
           return order2.compareTo(order1);
        }
    }

    public String largestNumber(int[] nums) {
        // Get input integers as strings.
        String[] asStrs = new String[nums.length];
        for (int i = 0; i < nums.length; i++) {
            asStrs[i] = String.valueOf(nums[i]);
        }

        // Sort strings according to custom comparator.
        Arrays.sort(asStrs, new LargerNumberComparator());

        // If, after being sorted, the largest number is `0`, the entire number
        // is zero.
        if (asStrs[0].equals("0")) {
            return "0";
        }

        // Build largest number from sorted array.
        String largestNumberStr = new String();
        for (String numAsStr : asStrs) {
            largestNumberStr += numAsStr;
        }

        return largestNumberStr;
    }
}