Determine if an Integer's Square Root Is an Integer in Java Article by Abdallah Sawan
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@ -35,7 +35,7 @@ public class PerfectSquareUtil {
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return x1 == x2 && n % x1 == 0L;
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}
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public static boolean isSquareNumberWithOptimization(long n) {
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public static boolean isPerfectSquareWithOptimization(long n) {
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if (n < 0)
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return false;
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switch ((int) (n & 0xF)) {
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@ -12,7 +12,7 @@ public class PerfectSquareUnitTest {
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assertEquals(expectedValue, PerfectSquareUtil.isPerfectSquareByUsingSqrt(n));
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assertEquals(expectedValue, PerfectSquareUtil.isPerfectSquareByUsingBinarySearch(1, Integer.MAX_VALUE, n));
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assertEquals(expectedValue, PerfectSquareUtil.isPerfectSquareByUsingNewtonMethod(n));
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assertEquals(expectedValue, PerfectSquareUtil.isSquareNumberWithOptimization(n));
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assertEquals(expectedValue, PerfectSquareUtil.isPerfectSquareWithOptimization(n));
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}
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@Test
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@ -22,6 +22,6 @@ public class PerfectSquareUnitTest {
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assertEquals(expectedValue, PerfectSquareUtil.isPerfectSquareByUsingSqrt(n));
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assertEquals(expectedValue, PerfectSquareUtil.isPerfectSquareByUsingBinarySearch(1, Integer.MAX_VALUE, n));
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assertEquals(expectedValue, PerfectSquareUtil.isPerfectSquareByUsingNewtonMethod(n));
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assertEquals(expectedValue, PerfectSquareUtil.isSquareNumberWithOptimization(n));
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assertEquals(expectedValue, PerfectSquareUtil.isPerfectSquareWithOptimization(n));
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}
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}
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