LeetCode 29. Divide Two Integers

Divide two integers without using multiplication, division and mod operator.

If it is overflow, return MAX_INT.

没有使用移位，而是采取了小学竖式计算的原理

public int divide(int dividend, int divisor) {

long dividend_long = dividend;
long divisor_long = divisor;
int flag = 1;
if (divisor_long == 0) {
return Integer.MAX_VALUE;
}
if (divisor_long < 0) {
divisor_long = -divisor_long;
flag = flag * -1;
}
//分子转化成字符串
String fenzi = String.valueOf(dividend_long);
//处理负号
if (fenzi.charAt(0) == '-') {
fenzi = fenzi.substring(1);
flag = flag *  -1;
}
//设置当前的左右游标
int left = 0;
int right = 1;
int length = fenzi.length();
StringBuffer sb = new StringBuffer();
int yu = 0;
while (right <= length) {
long now = Integer.parseInt(fenzi.substring(left,right));//将当前要计算的字符串转化为数字
now = now + yu * 10;
long sum = 0;
long result_now = 0;
while ((sum + divisor_long) <= now) {
sum += divisor_long;
result_now++;
}
yu = (int)(now - sum);
if (result_now == 0) {
if (yu != 0) {
if (sb.length() != 0) {
sb.append(0);
}
left = right;
right++;

} else {
sb.append(0);
left = right;
right ++;
}

} else {
sb.append(result_now);
left = right;
right ++;
}
}
if (sb.length() == 0) {
return  0;
}
Long result =  flag * Long.parseLong(sb.toString());
if (result > Integer.MAX_VALUE) {
return Integer.MAX_VALUE;
} else if (result < Integer.MIN_VALUE) {
return Integer.MIN_VALUE;
} else {
return result.intValue();
}
}