LintCode:乘积最大子序列

LintCode:乘积最大子序列

动态规划，不过与和最大子序列不同，原因是乘积有正负，因而在动规数组中不能只记录最大值，还需要记录最小值，如：nums = [1,0,-1,2,3,-5,-2]，如果只用一个动规数组，dp=[1,1,1,2,6,6,6],即nums[3]与nums[4] 的积，显然是不正确的，因为nums[3]*nums[4]*nums[5]*nums[6]=60 。为了解决这一问题，我们需要再增加一个动规数组，用来记录乘积最小值，防止由于”负负得正“引起的正负值的问题。

Python

class Solution:
    # @param nums: an integer[]
    # @return: an integer
    def maxProduct(self, nums):
        # write your code here
        if len(nums) == 0:
            return 0
        max_ans = [nums[0] for i in range(len(nums))]
        min_ans = [nums[0] for i in range(len(nums))]
        for i in range(1, len(nums)):
            max_ans[i] = max(max_ans[i-1]*nums[i], min_ans[i-1]*nums[i], nums[i])
            min_ans[i] = min(max_ans[i-1]*nums[i], min_ans[i-1]*nums[i], nums[i])
        return max(max_ans)

Java

public class Solution {
    /** * @param nums: an array of integers * @return: an integer */
    public int maxProduct(int[] nums) {
       if(nums.length == 0){
            return 0;
        }
        int[] maxAns = new int[nums.length];
        int[] minAns = new int[nums.length];
        maxAns[0] = nums[0];
        minAns[0] = nums[0];

        for(int i=1; i<nums.length; i++){
            maxAns[i] = max(maxAns[i-1]*nums[i], minAns[i-1]*nums[i], nums[i]);
            minAns[i] = min(maxAns[i-1]*nums[i], minAns[i-1]*nums[i], nums[i]);
        }
        return max(maxAns);
    }

    private static int max(int[] List) {
        int maxNum = List[0];
        for(int i=0; i<List.length; i++){
            if(List[i] > maxNum){
                maxNum = List[i];
            }
        }
        return maxNum;
    }

    private static int max(int m, int n, int k) {
        int maxNum = m > n? m:n;
        maxNum = maxNum > k? maxNum: k;
        return maxNum;
    }

    private static int min(int m, int n, int k){
        int minNum = m < n? m : n;
        minNum = minNum < k? minNum: k;
        return minNum;
    }
}