[题目] 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
.
[分析] 注意到负负得正,所以求解过程中需要分别计算保留以当前数组元素A[i]结尾的连续子数组乘积的最大值和最小值
public class Solution { // O(N)空间 + O(N)时间 public int maxProduct1(int[] A) { if (A == null || A.length == 0) return 0; int n = A.length; int[] max = new int[n]; int[] min = new int[n]; // 初始化条件 max[0] = A[0]; min[0] = A[0]; int finalMax = A[0]; for (int i = 1; i < n; i++) { // 递推式 max[i] = Math.max(A[i], Math.max(max[i - 1] * A[i], min[i - 1] * A[i])); min[i] = Math.min(A[i], Math.min(max[i - 1] * A[i], min[i - 1] * A[i])); if (max[i] > finalMax) finalMax = max[i]; } return finalMax; } // O(1)空间 + O(N)时间,仅需保留上一步的结果 public int maxProduct(int[] A) { if (A == null || A.length == 0) return 0; int n = A.length; int[] max = new int[2]; int[] min = new int[2]; max[0] = A[0]; min[0] = A[0]; int finalMax = A[0]; for (int i = 1; i < n; i++) { max[1] = Math.max(A[i], Math.max(max[0] * A[i], min[0] * A[i])); min[1] = Math.min(A[i], Math.min(max[0] * A[i], min[0] * A[i])); if (max[1] > finalMax) finalMax = max[1]; max[0] = max[1]; min[0] = min[1]; } return finalMax; } }