66. Plus One
Given a non-negative number represented as an array of digits, plus one to the number.
The digits are stored such that the most significant digit is at the head of the list.
public class Solution {
public int[] plusOne(int[] digits) {
int len = digits.length;
int sum = 0;
int sum1 = 0;
int[] res= new int[len];
int[] res1= new int[len+1];
for(int i = 0; i< digits.length; i++){
sum +=Math.pow(10,len-i-1)*digits[i];
}
for(int i = 0; i<digits.length; i++){
sum1 += Math.pow(10,len-i-1)*1;
}
if(sum1*9==sum){
sum= sum+1;
res1[0]=1;
for(int j=1;j<res1.length;j++){
res1[j]=0;
}
return res1;
}else{
for(int j=0;j<res.length;j++){
res[res.length-j-1]=sum%10;
sum=sum/10;
}
return res;
}
}
}
数值小时可通过,但是没有考虑到数值的范围。待会再改:
一把辛酸泪…………………………
本文讨论了数值加一算法的实现细节,重点解决了数值范围内的边界处理问题,并提出了改进策略,确保算法在不同场景下都能准确执行加一操作。
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