modify code

Author: algorithmzuo

src/class26/Code01_SumOfSubarrayMinimums.java

@@ -1,11 +1,10 @@
 package class26;
 
-import java.util.Stack;
-
 // 测试链接：https://leetcode.com/problems/sum-of-subarray-minimums/
 // subArrayMinSum1是暴力解
 // subArrayMinSum2是最优解的思路
 // sumSubarrayMins是最优解思路下的单调栈优化
+// Leetcode上不要提交subArrayMinSum1、subArrayMinSum2方法，因为没有考虑取摸
 // Leetcode上只提交sumSubarrayMins方法，时间复杂度O(N)，可以直接通过
 public class Code01_SumOfSubarrayMinimums {
 
@@ -71,8 +70,9 @@ public static int[] rightNearLess2(int[] arr) {
 }
 
 public static int sumSubarrayMins(int[] arr) {
-int[] left = nearLessEqualLeft(arr);
-int[] right = nearLessRight(arr);
+int[] stack = new int[arr.length];
+int[] left = nearLessEqualLeft(arr, stack);
+int[] right = nearLessRight(arr, stack);
 long ans = 0;
 for (int i = 0; i < arr.length; i++) {
 long start = i - left[i];
@@ -83,34 +83,34 @@ public static int sumSubarrayMins(int[] arr) {
 return (int) ans;
 }
 
-public static int[] nearLessEqualLeft(int[] arr) {
+public static int[] nearLessEqualLeft(int[] arr, int[] stack) {
 int N = arr.length;
 int[] left = new int[N];
-Stack<Integer> stack = new Stack<>();
+int size = 0;
 for (int i = N - 1; i >= 0; i--) {
-while (!stack.isEmpty() && arr[i] <= arr[stack.peek()]) {
-left[stack.pop()] = i;
+while (size != 0 && arr[i] <= arr[stack[size - 1]]) {
+left[stack[--size]] = i;
 }
-stack.push(i);
+stack[size++] = i;
 }
-while (!stack.isEmpty()) {
-left[stack.pop()] = -1;
+while (size != 0) {
+left[stack[--size]] = -1;
 }
 return left;
 }
 
-public static int[] nearLessRight(int[] arr) {
+public static int[] nearLessRight(int[] arr, int[] stack) {
 int N = arr.length;
 int[] right = new int[N];
-Stack<Integer> stack = new Stack<>();
+int size = 0;
 for (int i = 0; i < N; i++) {
-while (!stack.isEmpty() && arr[stack.peek()] > arr[i]) {
-right[stack.pop()] = i;
+while (size != 0 && arr[stack[size - 1]] > arr[i]) {
+right[stack[--size]] = i;
 }
-stack.push(i);
+stack[size++] = i;
 }
-while (!stack.isEmpty()) {
-right[stack.pop()] = N;
+while (size != 0) {
+right[stack[--size]] = N;
 }
 return right;
 }