278. First Bad Version

278. First Bad Version

Question

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.

Example:

Given n = 5, and version = 4 is the first bad version.

call isBadVersion(3) -> false
call isBadVersion(5) -> true
call isBadVersion(4) -> true

Then 4 is the first bad version.

Attention:

• 在求mid时， (low + high) / 2会造成溢出。
• 使用 low + (high - low) / 2最为稳妥。

Thinking:

• Method 1: 递归， 二分法

/* The isBadVersion API is defined in the parent class VersionControl.
      boolean isBadVersion(int version); */

public class Solution extends VersionControl {
    public int firstBadVersion(int n) {
        if(n == 1) return n;
        int low = 1, high = n;
        return binarySearch(low, high);
    }
    private int binarySearch(int low, int high){
        if(low >= high) return low;
        int mid = low + (high - low ) / 2;
        boolean cur = isBadVersion(mid);
        boolean pre = isBadVersion(mid - 1);
        if(cur && !pre){
            return mid;
        }else if(cur && pre){
            return binarySearch(low, mid - 1);
        }else{
            return binarySearch(mid + 1, high);
        }
    }
}

• Method 2:循环 二分法

/* The isBadVersion API is defined in the parent class VersionControl.
      boolean isBadVersion(int version); */

public class Solution extends VersionControl {
    public int firstBadVersion(int n) {
        if(n == 0 || n == 1) return n;
        int low = 1, high = n;
        int mid = 0;
        while(low < high){
            mid = low + (high - low) / 2;
            if(isBadVersion(mid))
                high = mid;
            else
                low = mid + 1;
        }
        return low;
    }
}

Second time

1. Binary search, we need to consider current result and previoud result. ```Java /* The isBadVersion API is defined in the parent class VersionControl.
  boolean isBadVersion(int version); */ public class Solution extends VersionControl {
public int firstBadVersion(int n) {
    int low = 1, high = n;
    return getResult(1, n + 1);
}
public int getResult(int low, int high){
    int mid = (high - low) / 2 + low;
    boolean res = isBadVersion(mid);
    if(mid - 1 < 1 && res) return 1;
    boolean pre = isBadVersion(mid - 1);
    if(res && !pre) return mid;
    else if(res && pre) return getResult(low, mid);
    else return getResult(mid, high);
} } ```