forked from Blankj/awesome-java-leetcode
/
Sqrt.java
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Sqrt.java
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public class Sqrt {
/*
Sqrt(x)
Description
Implement int sqrt(int x).
Compute and return the square root of x, where x is guaranteed to be a non-negative integer.
Since the return type is an integer, the decimal digits are truncated and only the integer part of the result is returned.
Example 1:
Input: 4
Output: 2
Example 2:
Input: 8
Output: 2
Explanation: The square root of 8 is 2.82842..., and since
the decimal part is truncated, 2 is returned.
Tags: Binary Search, Math
*/
/*
思路
题意是求平方根,参考 牛顿迭代法求平方根,然后再参考维基百科的 Integer square root 即可。
*/
public static int mySqrt(int x) {
if (x < 1) {
return 0;
}
long y = x;
while (y * y > x) {
y = (long)(0.5 * (y + x / y));
}
return (int)y;
}
public static int mySqrtWithMid(int x) {
if(x < 1) {
return 0;
}
int left = 0;
int right = x;
while(true) {
int mid = left + ((right - left) >> 1);
if (mid < 1) {
return 1;
}
if (mid > x / mid) {
right = mid;
} else {
if (mid + 1 > x/(mid + 1)) {
return mid;
}
left = mid + 1;
}
}
}
public static void main(String[] args) {
int input1 = 4;
System.out.println("input: " +input1+ " sqrt: " + mySqrt(input1));
System.out.println("input: " +input1+ " mySqrtWithMid: " + mySqrtWithMid(input1));
int input2 = 8;
System.out.println("input: " +input2+ " sqrt: " + mySqrt(input2));
System.out.println("input: " +input2+ " mySqrtWithMid: " + mySqrtWithMid(input2));
}
}