diff --git a/src/class25/Code02_AllTimesMinToMax.java b/src/class25/Code02_AllTimesMinToMax.java
index e574b8b..2f26ac1 100644
--- a/src/class25/Code02_AllTimesMinToMax.java
+++ b/src/class25/Code02_AllTimesMinToMax.java
@@ -67,7 +67,8 @@ public class Code02_AllTimesMinToMax {
 	// 本题可以在leetcode上找到原题
 	// 测试链接 : https://leetcode.com/problems/maximum-subarray-min-product/
 	// 注意测试题目数量大，要取模，但是思路和课上讲的是完全一样的
-	// 注意溢出的处理即可
+	// 注意溢出的处理即可，也就是用long类型来表示累加和
+	// 还有优化就是，你可以用自己手写的数组栈，来替代系统实现的栈，也会快很多
 	public static int maxSumMinProduct(int[] arr) {
 		int size = arr.length;
 		long[] sums = new long[size];
@@ -76,17 +77,20 @@ public class Code02_AllTimesMinToMax {
 			sums[i] = sums[i - 1] + arr[i];
 		}
 		long max = Long.MIN_VALUE;
-		Stack<Integer> stack = new Stack<Integer>();
+		int[] stack = new int[size];
+		int stackSize = 0;
 		for (int i = 0; i < size; i++) {
-			while (!stack.isEmpty() && arr[stack.peek()] >= arr[i]) {
-				int j = stack.pop();
-				max = Math.max(max, (stack.isEmpty() ? sums[i - 1] : (sums[i - 1] - sums[stack.peek()])) * arr[j]);
+			while (stackSize != 0 && arr[stack[stackSize - 1]] >= arr[i]) {
+				int j = stack[--stackSize];
+				max = Math.max(max,
+						(stackSize == 0 ? sums[i - 1] : (sums[i - 1] - sums[stack[stackSize - 1]])) * arr[j]);
 			}
-			stack.push(i);
+			stack[stackSize++] = i;
 		}
-		while (!stack.isEmpty()) {
-			int j = stack.pop();
-			max = Math.max(max, (stack.isEmpty() ? sums[size - 1] : (sums[size - 1] - sums[stack.peek()])) * arr[j]);
+		while (stackSize != 0) {
+			int j = stack[--stackSize];
+			max = Math.max(max,
+					(stackSize == 0 ? sums[size - 1] : (sums[size - 1] - sums[stack[stackSize - 1]])) * arr[j]);
 		}
 		return (int) (max % 1000000007);
 	}