solutions

Author: liuhanqu

algorithms/easy/fizzBuzz/fizzBuzz.js

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+// description: https://leetcode-cn.com/explore/interview/card/top-interview-questions-easy/25/math/60/
+// Fizz Buzz
+
+/**
+ * @param {number} n
+ * @return {string[]}
+ */
+var fizzBuzz = function(n) {
+  var ret = [];
+  var tmp = "";
+  for (var i = 1; i <= n; i++) {
+    tmp = "" + i;
+    if (i % 3 === 0) {
+      tmp = "Fizz";
+    } else if (i % 5 === 0) {
+      tmp = "Buzz";
+    }
+    if (i % 3 === 0 && i % 5 === 0) {
+      tmp = "FizzBuzz";
+    }
+    ret.push(tmp);
+  }
+  return ret;
+};

algorithms/easy/solution/solution.js

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+// description: https://leetcode-cn.com/explore/interview/card/top-interview-questions-easy/8/sorting-and-searching/53/
+// 第一个错误的版本
+
+/**
+ * Definition for isBadVersion()
+ *
+ * @param {integer} version number
+ * @return {boolean} whether the version is bad
+ * isBadVersion = function(version) {
+ *     ...
+ * };
+ */
+
+/**
+ * @param {function} isBadVersion()
+ * @return {function}
+ */
+var solution = function(isBadVersion) {
+  /**
+   * @param {integer} n Total versions
+   * @return {integer} The first bad version
+   */
+  return function(n) {
+    var low = 0;
+    var high = n - 1;
+    var mid;
+    while (low < high) {
+      mid = Math.floor((high - low) / 2) + low;
+      if (isBadVersion(mid)) {
+        high = mid - 1;
+      } else {
+        low = mid + 1;
+      }
+    }
+    return isBadVersion(low) ? low : low + 1;
+  };
+};