feat: add typescript solution to lc problem: No.1567.Maximum Length of Subarray With Positive Product (doocs#513)

zhaocchen · web-flow · commit b98eb258fc94 · 2021-07-15T17:48:57.000+08:00
diff --git a/solution/1500-1599/1567.Maximum Length of Subarray With Positive Product/README.md b/solution/1500-1599/1567.Maximum Length of Subarray With Positive Product/README.md
@@ -124,6 +124,31 @@ class Solution {
 }
 ```
 
+### **TypeScript**
+
+```ts
+function getMaxLen(nums: number[]): number {
+    // 连续正数计数n1, 连续负数计数n2
+    let n1 = nums[0] > 0 ? 1 : 0, n2 = nums[0] < 0 ? 1 : 0;
+    let ans = n1;
+    for (let i = 1; i < nums.length; ++i) {
+        let cur = nums[i];
+        if (cur == 0) {
+            n1 = 0, n2 = 0;
+        } else if (cur > 0) {
+            ++n1;
+            n2 = n2 > 0 ? n2 + 1 : 0;
+        } else {
+            let t1 = n1, t2 = n2;
+            n1 = t2 > 0 ? t2 + 1 : 0;
+            n2 = t1 + 1;
+        }
+        ans = Math.max(ans, n1);
+    }
+    return ans;
+};
+```
+
 ### **C++**
 
 ```cpp
diff --git a/solution/1500-1599/1567.Maximum Length of Subarray With Positive Product/README_EN.md b/solution/1500-1599/1567.Maximum Length of Subarray With Positive Product/README_EN.md
@@ -119,6 +119,31 @@ class Solution {
 }
 ```
 
+### **TypeScript**
+
+```ts
+function getMaxLen(nums: number[]): number {
+    // 连续正数计数n1, 连续负数计数n2
+    let n1 = nums[0] > 0 ? 1 : 0, n2 = nums[0] < 0 ? 1 : 0;
+    let ans = n1;
+    for (let i = 1; i < nums.length; ++i) {
+        let cur = nums[i];
+        if (cur == 0) {
+            n1 = 0, n2 = 0;
+        } else if (cur > 0) {
+            ++n1;
+            n2 = n2 > 0 ? n2 + 1 : 0;
+        } else {
+            let t1 = n1, t2 = n2;
+            n1 = t2 > 0 ? t2 + 1 : 0;
+            n2 = t1 + 1;
+        }
+        ans = Math.max(ans, n1);
+    }
+    return ans;
+};
+```
+
 ### **C++**
 
 ```cpp
diff --git a/solution/1500-1599/1567.Maximum Length of Subarray With Positive Product/Solution.ts b/solution/1500-1599/1567.Maximum Length of Subarray With Positive Product/Solution.ts
@@ -0,0 +1,20 @@
+function getMaxLen(nums: number[]): number {
+    // 连续正数计数n1, 连续负数计数n2
+    let n1 = nums[0] > 0 ? 1 : 0, n2 = nums[0] < 0 ? 1 : 0;
+    let ans = n1;
+    for (let i = 1; i < nums.length; ++i) {
+        let cur = nums[i];
+        if (cur == 0) {
+            n1 = 0, n2 = 0;
+        } else if (cur > 0) {
+            ++n1;
+            n2 = n2 > 0 ? n2 + 1 : 0;
+        } else {
+            let t1 = n1, t2 = n2;
+            n1 = t2 > 0 ? t2 + 1 : 0;
+            n2 = t1 + 1;
+        }
+        ans = Math.max(ans, n1);
+    }
+    return ans;
+};
