(Java)LeetCode-29. Divide Two Integers
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Divide two integers without using multiplication, division and mod operator.
If it is overflow, return MAX_INT.
刚看这一题时确实没什么思路,还想着一直减减减,后来看到tags里有二分查找,就有思路了。amazing
比如说 25/3 这一题,用一个Map存上(3-1)(6-2)(12-4)(24-8),因为下一个48大于了25,所以就结束了,然后从后向前遍历这个Map,这里是将其转换为list实现遍历的
只要找到比25小的,就与key做差,然后结果加上value,在这里,首先25-24 = 1,result = 0 +8 =8;由于1<除数了,所以结束了,答案就是8
难点在于溢出的处理上,因为int 可以取到的值为 -2147483648 到 2147483647,所以我将所有情况都变成两个负数相除了,只有一个特殊情况是-2147483648/1
我看了其他人的想法,虽然不能用乘除,但是他们用了《》位操作运算符,好机智,速度还快,比我厉害。代码如下:
package lzz;import java.util.ArrayList;import java.util.LinkedHashMap;import java.util.List;import java.util.Map;public class Solution {public int divide(int dividend, int divisor) {if(divisor == 0){return Integer.MAX_VALUE;}if(divisor == 1){ return dividend;}if(dividend < 0 && divisor <0){return divideTwoMinus(dividend,divisor);}else if(dividend > 0 && divisor <0){return -divideTwoMinus(-dividend,divisor);}else if(dividend < 0 && divisor > 0){return -divideTwoMinus(dividend,-divisor);}else{return divideTwoMinus(-dividend,-divisor);} }private int divideTwoMinus(int dividend, int divisor){Map<Integer,Integer> map = new LinkedHashMap<Integer,Integer>();int temp = divisor;int dif = 1;int num = 0;while(temp >= dividend){map.put(temp, dif);temp = temp + temp;if(temp >= 0)break;dif = dif + dif;}List<Map.Entry<Integer,Integer>> list = new ArrayList<Map.Entry<Integer,Integer>>(map.entrySet());int result = 0;num = list.size();while(dividend <= divisor){if(list.get(num-1).getKey()>=dividend){result += list.get(num-1).getValue();dividend = dividend - list.get(num-1).getKey();}num --;}if(result>=0)return result;elsereturn Integer.MAX_VALUE;} public static void main(String[] args){ Solution sol = new Solution(); System.out.print(sol.divide(1038925803,-2147483648)); }}
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