1945. 字符串转化后的各位数字之和

Author: w2xi

README.md

@@ -174,6 +174,7 @@
 |1047|[删除字符串中的所有相邻重复项](https://leetcode.cn/problems/remove-all-adjacent-duplicates-in-string/)|[JavaScript](./algorithms/remove-all-adjacent-duplicates-in-string.js)|Easy|
 |1780|[判断一个数字是否可以表示成三的幂的和](https://leetcode.cn/problems/check-if-number-is-a-sum-of-powers-of-three/)|[JavaScript](./algorithms/check-if-number-is-a-sum-of-powers-of-three.js)|Medium|
 |1832|[判断句子是否为全字母句](https://leetcode.cn/problems/check-if-the-sentence-is-pangram/)|[JavaScript](./algorithms/check-if-the-sentence-is-pangram.js)|Easy|
+|1945|[字符串转化后的各位数字之和](https://leetcode.cn/problems/sum-of-digits-of-string-after-convert/)|[JavaScript](./algorithms/sum-of-digits-of-string-after-convert.js)|Easy|
 |面试题 04.12|[面试题 04.12. 求和路径](https://leetcode.cn/problems/paths-with-sum-lcci/)|[JavaScript](./algorithms/paths-with-sum-lcci.js)|Medium|
 |面试题 02.07|[面试题 02.07. 链表相交](https://leetcode.cn/problems/intersection-of-two-linked-lists-lcci/)|[JavaScript](./algorithms/intersection-of-two-linked-lists-lcci.js)|Easy|
 |剑指 Offer 05. 替换空格|[剑指 Offer 05. 替换空格](https://leetcode.cn/problems/ti-huan-kong-ge-lcof/)|[JavaScript](./algorithms/ti-huan-kong-ge-lcof.js)|Easy|

algorithms/sum-of-digits-of-string-after-convert.js

@@ -0,0 +1,23 @@
+/**
+ * 1945. 字符串转化后的各位数字之和
+ * @param {string} s
+ * @param {number} k
+ * @return {number}
+ */
+ var getLucky = function(s, k) {
+  let result = 0;
+  let sum = '';
+
+  for (let i = 0; i < s.length; i++) {
+    sum += s[i].charCodeAt() - 97 + 1;
+  }
+  while (k--) {
+    result = 0;
+    for (let i = 0; i < sum.length; i++) {
+      result += Number(sum[i]);
+    }
+    sum = result.toString();
+  }
+
+  return result;
+};

vs-lt/1945.字符串转化后的各位数字之和.js

@@ -0,0 +1,31 @@
+/*
+ * @lc app=leetcode.cn id=1945 lang=javascript
+ *
+ * [1945] 字符串转化后的各位数字之和
+ */
+
+// @lc code=start
+/**
+ * @param {string} s
+ * @param {number} k
+ * @return {number}
+ */
+var getLucky = function(s, k) {
+  let result = 0;
+  let sum = '';
+
+  for (let i = 0; i < s.length; i++) {
+    sum += s[i].charCodeAt() - 97 + 1;
+  }
+  while (k--) {
+    result = 0;
+    for (let i = 0; i < sum.length; i++) {
+      result += Number(sum[i]);
+    }
+    sum = result.toString();
+  }
+
+  return result;
+};
+// @lc code=end
+