Programmers set,intersection union, leetcode splice

Programmers/[1차] 뉴스 클러스터링.py

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+import re
+
+
+def solution(str1, str2):
+    answer = 0
+    str1 = str1.upper()
+    str2 = str2.upper()
+    str1_list = []
+    str2_list = []
+
+    for i in range(len(str1) - 1):
+        temp = str1[i:i + 2]
+        if temp.isalpha() == True:
+            str1_list.append(temp)
+
+    for i in range(len(str2) - 1):
+        temp = str2[i:i + 2]
+        if temp.isalpha() == True:
+            str2_list.append(temp)
+
+    intersection = set(str1_list) & set(str2_list)
+    union = set(str1_list) | set(str2_list)
+
+    top = 0
+    bot = 0
+    for key in intersection:
+        top += min(str1_list.count(key), str2_list.count(key))
+    for key in union:
+        bot += max(str1_list.count(key), str2_list.count(key))
+
+    if bot == 0:
+        answer = 1
+    else:
+        answer = top / bot
+    answer = int(answer * 65536)
+    return answer

leetcode JS/26. Remove Duplicates from Sorted Array.js

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+/**
+ * @param {number[]} nums
+ * @return {number}
+ */
+var removeDuplicates = function (nums) {
+  let i = 0;
+  while (i < nums.length - 1) {
+    j = i + 1;
+    if (nums[i] === nums[j]) {
+      nums.splice(j, 1);
+    } else {
+      i++;
+    }
+  }
+};