69. Sqrt(x)
题目
Implement int sqrt(int x).
Compute and return the square root of x.
x is guaranteed to be a non-negative integer.
Example 1:
Input: 4
Output: 2
Example 2:
Input: 8
Output: 2
Explanation: The square root of 8 is 2.82842..., and since we want to return an integer, the decimal part will be trunca
解析
class Solution_69 {
public:
int mySqrt(int x) {
int ret = 0;
int mx = INT_MAX;
while (1)
{
long long temp = ret*ret;
long long temp1 = (long long)(ret+1)*(long long)(ret+1); //关于数据的题考虑越界和边界条件；*操作之前强制类型转换
if (temp1>mx)
{
return ret;
}
if (temp<= x && temp1>x) //考虑越界的问题
{
break;
}
ret++;
}
return ret;
}
int sqrt(int x) //牛顿逼近法
{
long r = x;
while (r*r > x)
r = (r + x / r) / 2;
return r;
}
//特比特别要注意两点：第一right要取x / 2 + 1 这个还不是最重要的，其实只是影响速度
//第二：要用x / middle > middle 来表示x > middle*middle 不然会溢出
//第三：判断相等时用x / middle >= middle && x / (middle + 1) < (middle + 1)
int sqrt_(int x)
{
if (x<2)
{
return x;
}
int l = 1, r = x/2+1;
int mid=0;
while (l<=r)
{
//mid = l + (r - l) / 2;
mid = l + ((r - l) >> 1); //位运算的优先级低于算术运算 bug: mid=l+(r-l)>>1
if (mid != 0)
{
if (x / (mid + 1)<mid + 1 && x / mid>=mid)
{
return mid;
}else if (x / mid>=mid) //不使用x>mid*mid
{
l = mid + 1;
}
else if(x / mid< mid)
{
r = mid - 1;
}
}
}
return mid;
}
int sqrt_ref(int x) {
if (x == 0){
return 0;
}
if (x < 0){
return -1;
}
int left = 1, right = x / 2 + 1, middle;
while (left <= right){
middle = (left + right) / 2;
if (x / middle >= middle && x / (middle + 1)<(middle + 1)){
return middle;
}
else if (x / middle>middle){
left = middle + 1;
}
else{
right = middle - 1;
}
}
return right;
}
};
题目来源