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algorithm
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src
/
arrays
/
array_14.java
algorithm
/
src
/
arrays
/
array_14.java
array_14.java 1.46 KB
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Gavery authored 2022年08月01日 22:03 +08:00 . review
package arrays;
/**题目:求连续子序列的最大乘积
* @Author Gavin
* @date 2022年01月09日 15:10
*/
public class array_14 {
/**
* 解题思路:
* 要求连续子序列的最大乘积,我们需要知道一点就是一个最小负数乘以一个负数就可能
* 为一个最大的正数,所以我们需要把最小的负数也保存起来
*/
public static int solution(int[] nums){
if(nums==null||nums.length==0)return 0;
int max=nums[0],curMax=nums[0],curMin=nums[0];
for (int i = 1; i < nums.length; i++) {
//核心点就是连续乘以每一个元素,要么是最大,要么就是最小
//curMax表示当前正在处理的最大值,核心点,当连续正数乘积之后遇到负数后的第一个正数
//curMax值就是这个正数的值,相当于从负数之后重新开始计算最大连续子序列乘积了
//同理,负数也一样
int a=curMax*nums[i],b=curMin*nums[i],c=nums[i];
curMax=max(a,b,c);
curMin=min(a,b,c);
max=Math.max(max,curMax);
}
return max;
}
public static int max(int a,int b,int c){
return Math.max(Math.max(a,b),c);
}
public static int min(int a,int b,int c){
return Math.min(Math.min(a,b),c);
}
public static void main(String[] args) {
int[] nums=new int[]{2,0,-1,3,4};
System.out.println(solution(nums));
}
}
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