leetcode-Sqrt(x)与Bad version

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本文介绍了二分查找算法在两个典型问题中的应用：计算平方根的整数部分和寻找首个故障版本。对于计算平方根问题，通过二分查找快速定位到最接近的整数值；在寻找首个故障版本的问题中，则利用二分查找高效地从一系列版本中找出首个故障版本。

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Sqrt(x) vs First bad version

1. Sqrt(x)

题目：Implement

int
 sqrt(int x)

思路：看到这道题目首先注意到返回的值都是整型变量。那么当x小于第一个i^2（i>=2）的时候，sqrt(x)的值都是i-1，于是题目就转换成为求第一个x/i^2==0的i的问题。我们可以用暴力搜索，直接遍历i，但是这样会造成TEL。所以我们可以使用二分法搜索。

类比：这道题目和求Bad version的题目很像。参考2.

注意点：

(1) 注意二分搜索中的整型下标的范围，不要越界

(2) 注意x/i^2的写法，写成i^2有可能会越界，所以可以写成x/i/i

      int mySqrt(int x) {
        if (x <= 0)
        {
            return 0;
        }
        if ( x == 1)
        {
            return 1;
        }
        int low = 1, high = x, mid;
        mid = low + (high - low) / 2; //防止下标越界
        while ( low < high )
        {
            if ( x / mid / mid  ) //防止i^2越界
            {
                low = mid + 1;
            }
            else
            {
                high = mid;
            }
            mid = low + (high - low) / 2;
        }
        return mid-1;
      }

2. First bad version
题目：

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.

int firstBadVersion(int n) {
        int p = 1, r = n;
        int mid = p+(r-p)/2;
        while ( p<r ){
            if ( isBadVersion(mid) )
            {
                r = mid;
            } //is Bad 
            else{
                p = mid+1;
            } //is good 
            mid = p+(r-p)/2;
        }
        return mid;
    }