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
.
minLast[i] 表示以A[i]结尾的最小值,maxLast[i]表示以A[i]结尾的最大值。
minLast[i] = Math.min(A[i], Math.min(minLast[i - 1]*A[i], maxLast[i - 1]*A[i]));
maxLast[i] = Math.max(A[i], Math.max(maxLast[i - 1]*A[i], minLast[i - 1]*A[i]));
1 public class Solution { 2 public int maxProduct(int[] A) { 3 int length = A.length; 4 if (length == 0) { 5 return 0; 6 } 7 int[] minLast = new int[length]; 8 int[] maxLast = new int[length]; 9 minLast[0] = A[0]; 10 maxLast[0] = A[0]; 11 int res = Integer.MIN_VALUE; 12 for (int i = 1; i <length ; i++) { 13 minLast[i] = Math.min(A[i], Math.min(minLast[i - 1]*A[i], maxLast[i - 1]*A[i])); 14 maxLast[i] = Math.max(A[i], Math.max(maxLast[i - 1]*A[i], minLast[i - 1]*A[i])); 15 } 16 for (int i : maxLast) { 17 if (i>res) res = i; 18 } 19 return res; 20 } 21 }