278第一个错误的版本.md

这道题是二分法查找第一个符合条件的元素

需要注意两个边界条件

1 start = mid + 1 : 如果用start = mid,只剩两个相邻的元素时候可能会卡死 2 mid = start + (end - start)/2 : 如果用(end + start)/2, end + start可能会超过整数最大值

/* 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 start = 1;
        int end = n;
        int mid = start + (end - start)/2;
        //（low+high)有超出int上限的可能性
        while(start < end){
            mid = start + (end - start)/2;
            if(isBadVersion(mid)){
                end = mid;
            }else{
                start = mid + 1;
            }
        }
        return end;
    }
}