Comments (5)
还有一种解法,先从最后一位加1并往前进位循环一次,然后判断数组首位是否大于等于10 然后确定是否要进位
public int[] plusOne(int[] digits) {
int length = digits.length;
digits[length -1] = digits[length-1] + 1;
while(length > 1){
int tmp = digits[length-1];
if(tmp >= 10){
int divide = tmp /10;
int mod = tmp % 10;
digits[length-1] = mod;
digits[length -2] = digits[length-2] + divide;
}
length --;
}
int tmp = digits[0];
if(tmp >= 10){
int[] di = new int[digits.length+1];
int divide = tmp /10;
int mod = tmp % 10;
digits[0] = mod ;
di[0] = divide;
for(int i = 0; i < digits.length; i++){
di[i+1] = digits[i];
}
return di;
}
return digits;
}
from algorithm007-class02.
public int[] plusOne(int[] digits) {
int length = digits.length;
int pos = length - 1;
digits[pos]++;
while(pos > 0){ //最高位涉及增加位数,后面单独处理
// 因为只加1,所以需要进位的情况为:当前位的值为10,则高位数字+1,当前位的值置0
if(digits[pos] >= 10){
int digits[pos] = 0;
int digits[pos-1] ++;
}
length --;
}
if(digits[0] >= 10){
int[] di = new int[digits.length+1];
digits[0] = 0 ; // 次高位
di[0] = 1; // 最高位
for(int i = 0; i < digits.length; i++){
di[i+1] = digits[i];
}
return di;
}
return digits;
}
没学过java,直接拷贝了@jeff-liu14 的代码,然后进行了一些修改。
思路是,我们已经知道增加的数值为1,所以取余的操作直接变成置0,进位操作直接变成高位加1.
from algorithm007-class02.
这个题我最开始还陷入了无趣,如果最后一位+1后大于9,就把数组的值转化为int数来+1,然后再转回数组。
这个方法会遇到到测试用例中数组特别大,超出了int的值域,导致提交失败的情况。
from algorithm007-class02.
这个问题解决如果一个正整数加一的角度考虑去实现问题,无法绕开的问题是数值的Max和Min值问题。不建议采用。
我把这个问题看成一个逢十进一的问题。数组高位开始检索满足十进一条件将数值赋值为0开始继续检索第二高位,如果不满足条件数值加1返回即可。以此类推代码如下:
public static int[] plusOne(int[] digits) {
int size = digits.length;
for (int j = size -1 ; j >= 0; j --) {
if(digits[j] == 9){
if(j == 0){
int[] r = new int[size + 1];
r[0] = 1;
return r;
}
digits[j] = 0;
}else{
digits[j] ++;
break;
}
}
return digits;
}
from algorithm007-class02.
学习了!
from algorithm007-class02.
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from algorithm007-class02.