LeetCode: Sqrt(x)
2014-06-27 08:26
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思路:二分查找,但是注意到一个INT数的平方会超过INT的表示范围,所以,中间过程需要用long long 类型保存。
code:
code:
class Solution { public: int sqrt(int x) { long long left = 0, right = x; while(left <= right){ long long mid = (left + right) / 2; if(mid*mid == x)return mid; else if(mid * mid < x) left = mid+1; else right = mid-1; } return right; } };
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