給予一個非負數的數字陣列，利用陣列內的元素排列出一個最大的數字，回傳值為字串
Test Case
這邊的解法直接丟上去 leetCode 還是會有問題，因為有些 testCase 沒覆蓋到，後續等有空時再繼續做…
| Numbers | Expected |
|---|---|
| [0, 0] | 0 |
| [10, 2] | 210 |
| [3, 30, 34, 5, 9] | 9534330 |
| [121, 12] | 12121 |
| [824, 938, 1399, 5607, 6973, 5703, 9609, 4398, 8247] | 9609938824824769735703560743981399 |
解題方法很多，重點的比較規則是將兩個數字字串合併後，再去比較，c-sharp的部分利用string.Compare()比較字串避免溢位
Solution in Js
class Solution {
constructor() {}
GetLargestNumber(numbers) {
let sortList = numbers.sort(function(x, y) {
let xFirst = x.toString() + y.toString();
let yFirst = y.toString() + x.toString();
return yFirst - xFirst;
});
let result = sortList.reduce((a, b) => {
if (a == b) return a.toString();
return a.toString() + b.toString();
});
return result;
}
}
module.exports = Solution;
Solution in C-sharp
其實就直接將數字陣列利用 IComparer 實作比較的方法，一開始不需要弄成 Dictionary…. 不過也懶得再重寫了，大概就是這個意思
public class Solution20190510
{
public string LargestNumber(int[] nums)
{
var myList = GetSortedList(nums);
return GetLargestNumberByList(myList);
}
private string GetLargestNumberByList(IEnumerable<KeyValuePair<string, int>> myList)
{
var result = "";
foreach (var keyValuePair in myList)
{
result += keyValuePair.Key;
}
return result;
}
private IEnumerable<KeyValuePair<string, int>> GetSortedList(IEnumerable<int> nums)
{
var dict = new Dictionary<string, int>();
foreach (var num in nums)
{
if (dict.ContainsKey(num.ToString()))
{
continue;
}
dict.Add(num.ToString(), num);
}
var result = dict.ToList();
result.Sort(mySort);
return result;
}
private int mySort(KeyValuePair<string, int> x, KeyValuePair<string, int> y)
{
var xFirst = x.Key + y.Key;
var yFirst = y.Key + x.Key;
return string.Compare(yFirst, xFirst);
}
}