week11 mission maximum-product-subarray

Author: dev-jonghoonpark

maximum-product-subarray/dev-jonghoonpark.md

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+- 문제: https://leetcode.com/problems/maximum-product-subarray/
+- 풀이: https://algorithm.jonghoonpark.com/2024/07/09/leetcode-152
+
+```java
+class Solution {
+    public int maxProduct(int[] nums) {
+        int max = Integer.MIN_VALUE;
+        int temp = 0;
+        int lastZeroIndex = 0;
+        for (int i = 0; i < nums.length; i++) {
+            int current = nums[i];
+            if (temp == 0) {
+                temp = current;
+                lastZeroIndex = i;
+            } else {
+                if (current == 0) {
+                    temp = 0;
+                } else if (temp > 0 && current < 0) {
+                    if (hasNextMinus(nums, i + 1)) {
+                        temp = temp * current;
+                    } else {
+                        temp = temp * current;
+                        for (int j = lastZeroIndex; j < i + 1; j++) {
+                            temp = temp / nums[j];
+                            if (temp > 0) {
+                                break;
+                            }
+                        }
+                    }
+                } else {
+                    temp = temp * current;
+                }
+            }
+            max = Math.max(max, temp);
+            if (temp < 0 && !hasNextMinus(nums, i + 1)) {
+                temp = 0;
+            }
+        }
+        return max;
+    }
+
+    private boolean hasNextMinus(int[] nums, int i) {
+        for (; i < nums.length; i++) {
+            if (nums[i] < 0) {
+                return true;
+            } else if (nums[i] == 0) {
+                return false;
+            }
+        }
+        return false;
+    }
+}
+```
+
+### TC, SC
+
+시간 복잡도는 O(n^2)이다. 공간 복잡도는 O(1)이다. hasNextMinus 이 적게 호출된다면 시간 복잡도는 O(n)에 가깝게 동작한다.