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yeahQing
java-leetcode
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java-leetcode
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jianzhioffer
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_2_001
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Solution.java
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37 lines (36 loc) · 1.46 KB
main
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java-leetcode
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jianzhioffer
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_2_001
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Solution.java
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package com.yeahqing.jianzhioffer._2_001;
/**
* @author YeahQing
* @date 2022/9/28
*/
class Solution {
/***
* 剑指 Offer II 001. 整数除法
* 给定两个整数 a 和 b ,求它们的除法的商 a/b ,要求不得使用乘号 '*'、除号 '/' 以及求余符号 '%'。
* 注意:
* 整数除法的结果应当截去(truncate)其小数部分,例如:truncate(8.345) = 8以及truncate(-2.7335) = -2
* 假设我们的环境只能存储 32 位有符号整数,其数值范围是 [−2^31,2^31−1]。本题中,如果除法结果溢出,则返回 2^31−1
* @param a 整数a
* @param b 整数b
* @return a/b
*/
public int divide(int a, int b) {
int MIN = Integer.MIN_VALUE, MAX = Integer.MAX_VALUE;
int MIN_LIMIT = MIN >> 1; // -1073741824
if(a == MIN && b == -1) return MAX; // 特判
boolean isPos = (a >= 0 || b <= 0) && (a <= 0 || b >= 0);
if(a > 0) a = -a;
if(b > 0) b = -b;
int ans = 0; // 最终的商
while(a <= b) {
int d = b, c = 1; // d为当前除数,c为当前商
while(d >= MIN_LIMIT && d + d >= a) { // 通过第一个条件防止d + d溢出
d += d; // 当前除数倍增,也可以用 d <<= 1;
c += c; // 当前商倍增,也可以用 c <<= 1;
}
a -= d; // a剩余部分
ans += c; // 累计当前商
}
return isPos ? ans : -ans;
}
}
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