228

Author: lzl124631x

leetcode/2188. Minimum Time to Finish the Race/README.md

@@ -72,7 +72,7 @@ public:
         int N = A.size(), len = 0;
         vector<long> best(numLaps, LONG_MAX), dp(numLaps + 1, INT_MAX);
         for (int i = 0; i < N; ++i) {
-            long f = A[i][0], r = A[i][1], sum = change, p = 1; // We assume we also need `change` time to use the first tire so that we don't need to care the first tire as a special case
+            long f = A[i][0], r = A[i][1], sum = change, p = 1; // We assume we also need `change` time to use the first tire so that we don't need to treat the first tire as a special case
             for (int j = 0; j < numLaps; ++j) {
                 sum += f * p;
                 if (f * p >= f + change) break; // If using the same tire takes no less time than changing the tire, stop further using the current tire

leetcode/228. Summary Ranges/README.md

@@ -33,34 +33,20 @@
 [8,9] --> "8->9"
 </pre>
 
-<p><strong>Example 3:</strong></p>
-
-<pre><strong>Input:</strong> nums = []
-<strong>Output:</strong> []
-</pre>
-
-<p><strong>Example 4:</strong></p>
-
-<pre><strong>Input:</strong> nums = [-1]
-<strong>Output:</strong> ["-1"]
-</pre>
-
-<p><strong>Example 5:</strong></p>
-
-<pre><strong>Input:</strong> nums = [0]
-<strong>Output:</strong> ["0"]
-</pre>
-
 <p>&nbsp;</p>
 <p><strong>Constraints:</strong></p>
 
 <ul>
 	<li><code>0 &lt;= nums.length &lt;= 20</code></li>
 	<li><code>-2<sup>31</sup> &lt;= nums[i] &lt;= 2<sup>31</sup> - 1</code></li>
 	<li>All the values of <code>nums</code> are <strong>unique</strong>.</li>
+	<li><code>nums</code> is sorted in ascending order.</li>
 </ul>
 
 
+**Companies**:  
+[Yandex](https://leetcode.com/company/yandex), [Facebook](https://leetcode.com/company/facebook), [Google](https://leetcode.com/company/google)
+
 **Related Topics**:  
 [Array](https://leetcode.com/tag/array/)
 
@@ -89,3 +75,25 @@ public:
     }
 };
 ```
+
+Or
+
+```cpp
+// OJ: https://leetcode.com/problems/summary-ranges/
+// Author: github.com/lzl124631x
+// Time: O(N)
+// Space: O(N)
+class Solution {
+public:
+    vector<string> summaryRanges(vector<int>& A) {
+        vector<string> ans;
+        int N = A.size();
+        for (int i = 0; i < N; ++i) {
+            int start = A[i];
+            while (i + 1 < N && A[i + 1] == A[i] + 1) ++i;
+            ans.push_back(to_string(start) + (A[i] == start ? "" : ("->" + to_string(A[i]))));
+        }
+        return ans;
+    }
+};
+```