modify code

Author: algorithmzuo

src/class25/Code02_AllTimesMinToMax.java

@@ -67,7 +67,8 @@ public static void main(String[] args) {
 // 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 static int maxSumMinProduct(int[] arr) {
 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);
 }