Sqrt(x)
# 题目
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.
来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/sqrtx
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。
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# 解答思路
使用二分法求其根，注意使用 x/res > res 来进行比较，如使用res * res >=x 有溢出风险。
需考虑代码健壮性排除0、1。
# 代码
package easy
import (
"testing"
)
/**
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.
来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/sqrtx
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。
*/
func mySqrt(x int) int {
if x <= 1 {
return x
}
min, max := 0, x
for max-min != 1 {
res := (min + max) / 2
if x/res >= res {
min = res
continue
}
max = res
}
return min
}
func TestMySqrt(t *testing.T) {
println(mySqrt(4))
println(mySqrt(8))
println(mySqrt(100))
}
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编辑 (opens new window)
上次更新: 2023/02/17, 16:53:03