You are a product manager and currently leading a team to develop a new product. Unfortunately, the latest version of your product fails the quality check. Since each version is developed based on the previous version, all the versions after a bad version are also bad.
Suppose you have n
versions [1, 2, ..., n]
and you want to find out the first bad one, which causes all the following ones to be bad.
You are given an API bool isBadVersion(version)
which will return whether version
is bad. Implement a function to find the first bad version. You should minimize the number of calls to the API.
Credits:Special thanks to @jianchao.li.fighter for adding this problem and creating all test cases.

這個(gè)題目很容易就想到使用二分法來(lái)搜索：
當(dāng)中點(diǎn)是Bad的時(shí)候，我們知道它后面就都是Bad，但是不知道這個(gè)是不是第一個(gè)bad，所以end=mid
當(dāng)中點(diǎn)是Good的時(shí)候，我們知道這個(gè)元素和這個(gè)元素之前的所有元素都是Good的，所以head=mid+1
至于什么時(shí)候終止，你可以用只剩兩個(gè)元素時(shí)的情況來(lái)測(cè)試一下。

/**
 * 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 head = 1;
        var end = n;
        while (head!==end) {
            var mid = parseInt(head + (end - head) / 2);
            if (isBadVersion(mid)) {
                end = mid;
            } else {
                head = mid+1;
            }
        }
        return head;
    };
};