[leetcode] 69 Sqrt(x)

来源：互联网发布：fetchassoc php 编辑：程序博客网时间：2024/05/22 00:23

问题描述：

Implement int sqrt(int x).

Compute and return the square root of x.

基本思路：

采用二分查找法；犹豫输入x是int型，int最大范围是2147483647 ，所以返回结果的范围0-46340. 注意要根据确定的区间去查找正确的返回值，如果大于46340的数的平方将超出int的表示范围。

代码：

int sqrt(int x) {        if(x==0||x==1)            return x;        if(x >=2147395600)            return 46340;        int high = 46339*2-1;        int low  = 1;                while(low <= high){            int mid = (high+low)/2;            int hpow = (mid+1)*(mid+1);            int lpow = mid*mid;            if(x>=lpow && x<hpow)                return mid;            else if(x<lpow)                high = mid-1;            else                 low = mid+1;        }            }

a better method

Since sqrt(x) is composed of binary bits, I calculate sqrt(x) by deciding every bit from the most significant to least significant.

 public int sqrt(int x) {    if(x==0)        return 0;    int h=0;    while((long)(1<<h)*(long)(1<<h)<=x) // firstly, find the most significant bit        h++;    h--;    int b=h-1;    int res=(1<<h);    while(b>=0){  // find the remaining bits        if((long)(res | (1<<b))*(long)(res |(1<<b))<=x)            res|=(1<<b);        b--;    }    return res;}

0 0