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/* * 69. Sqrt(x) * 2016-5-10 by Mingyang * Just binary search and logn complexity */ public static int sqrt(int x) { long i = 0; long j = x / 2 + 1; while (i <= j) { long mid = (i + j) / 2; if (mid * mid == x) return (int)mid; if (mid * mid < x) i = mid + 1; else j = mid - 1; } return (int)j; } }
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原文地址:http://www.cnblogs.com/zmyvszk/p/5480530.html