leetcode--Maximum Product Subarray

Find the contiguous subarray within an array (containing at least one number) which has the largest product.

For example, given the array

[2,3,-2,4]

,

the contiguous subarray

[2,3]

has the largest product =

6

.

题意：查询数组中有最大乘积的连续子数组。

分类：数组，动态规划

解法1：动态规划。由于乘积可能是负数，所以要保留两个数组，分别代表当前最大值和当前最小值。

递归规律：

max_arr[i] = Math.max(Math.max(max_arr[i-1]*nums[i],min_arr[i-1]*nums[i]),nums[i]);

min_arr[i] = Math.min(Math.min(max_arr[i-1]*nums[i],min_arr[i-1]*nums[i]),nums[i]);

public class Solution {
public int maxProduct(int[] nums) {
int len = nums.length;
int[] max_arr = new int[len];
int[] min_arr = new int[len];
int max = nums[0];
max_arr[0] = nums[0];
min_arr[0] = nums[0];
for(int i=1;i<len;i++){
max_arr[i] = Math.max(Math.max(max_arr[i-1]*nums[i],min_arr[i-1]*nums[i]),nums[i]);
min_arr[i] = Math.min(Math.min(max_arr[i-1]*nums[i],min_arr[i-1]*nums[i]),nums[i]);
max = Math.max(max,max_arr[i]);
}
return max;
}
}

优化一下代码，因为每次只要保留当前的最大值和最小值即可

public class Solution {
public int maxProduct(int[] nums) {
if(nums.length==0) return 0;
if(nums.length==1) return nums[0];
int curMax = nums[0];
int curMin = nums[0];
int res = nums[0];
for(int i=1;i<nums.length;i++){
int t = curMin*nums[i];
curMin = Math.min(nums[i], Math.min(curMax*nums[i], t));
curMax = Math.max(nums[i], Math.max(curMax*nums[i], t));
res = Math.max(res,curMax);
}
return res;
}
}