69. Sqrt(x)

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

• 69. Sqrt(x)