[LeetCode] 69. Sqrt(x)

標籤：output   ++   NPU   思想   想去   turned   個數   數字   ati   

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: 4Output: 2

Example 2:

Input: 8Output: 2Explanation: The square root of 8 is 2.82842..., and since              the decimal part is truncated, 2 is returned.


題意：開方
這裡肯定是不使用Math庫的前提下，
我們可以利用二分的思想去開方，開方反過來就是平方，Integer.MAX_VALUE 開方是46340，也就是最大的解，因為存在越界的情況
這個數字需要我們提前算出來。之後只要二分去找，mid*mid <= target , (mid+1)*(mid+1) > target 就行了

class Solution {    public int mySqrt(int x) {        int l = 0;        int r = 46340;        if (r*r < x)            return r;        if (x == 1 || x == 0)            return x;        while (r - l > 5) {            int mid = (l + r) / 2;            if (mid * mid < x) {                l = mid;            }            else if (mid * mid > x) {                r = mid;            }            else                return mid;        }        for (int i = l; i <= r; i++) {            if (i*i == x)                return i;            if (i*i < x && (i+1)*(i+1) > x)                return i;        }        return 0;    }}

 

[LeetCode] 69. Sqrt(x)