这是一道数值计算的题目,Code Ganker中指出数值计算类型题目通常就是二分法或者以2为基进行位处理。此题使用二分法,实现套路就是最经典的二分算法。数值计算题需要特别注意的就是数值越界处理。此题一开始没写对就是处理大数据时溢出,Method 1就是溢出版本,mid 为 int 类型,如果没有强转类型,则 mid * mid的结果仍然是int类型,赋给long类型的midSquare是不能防止数值溢出的。正如算平均数时使用减号避免加和溢出一样,可以使用除法来避免乘法溢出。
leetcode中其他数值计算题:pow(x,n),Divide Two Integers
[ref]
http://blog.csdn.net/linhuanmars/article/details/20089131
public class Solution { // Method 2 public int mySqrt(int x) { if (x <= 0) return 0; int left = 1, right = x / 2 + 1; int mid = 0; while (left <= right) { mid = left + ((right - left) >> 1); if (mid <= x / mid && (x / (mid + 1)) < mid + 1) { return mid; } else if (mid > x / mid) { right = mid - 1; } else { left = mid + 1; } } return 0; } // Method 1: fail @2147395599 public int mySqrt1(int x) { if (x <= 0) return 0; int left = 1, right = x / 2 + 1; int mid = 0; while (left <= right) { mid = left + ((right - left) >> 1); long midSquare = mid * mid; if (midSquare <= x && x < (mid + 1) * (mid + 1)) return mid; else if (midSquare > x) right = mid - 1; else left = mid + 1; } return right; } }