题目

解答

方案一

class Solution {public int mySqrt(int x) {if (x == 0) {return 0;}if (x < 4) {return 1;}for (int i = 2, length = Math.min(46342, x / 2 + 2); i < length; ++i) {int value = i * i;if (value == x) {return i;}if (value > x || value < 0) {return i - 1;}}return 0;}
}

方案二

class Solution {public int mySqrt(int x) {if (x == 0) {return 0;}if (x < 4) {return 1;}for (int i = Math.min(46340, x / 4 + 1); i > 1; --i) {long value = i * i;if (value == x) {return i;}if (value > 0 && value < x) {return i;}}return 0;}
}

方案三

class Solution {public int mySqrt(int x) {if (x == 0) {return 0;}if (x < 4) {return 1;}for (int length = Math.min(46342, x / 2 + 2); true;) {int value = length * length;if (value == x) {return length;}if (value < x && value > 0) {return length;}length = (length + x / length) / 2;}}
}

方案四

class Solution {public int mySqrt(int x) {if (x == 0) {return 0;}if (x < 4) {return 1;}for (int start = 1, end = Math.min(46342, x / 2 + 2); true;) {int mid = (start + end) / 2;if (mid == start || mid == end) {return mid;}int value = mid * mid;if (value == x) {return mid;}if (value < 0 || value > x) {end = mid;} else {start = mid;}}}
}

方案五

class Solution {public int mySqrt(int x) {if (x == 0) {return 0;}if (x < 4) {return 1;}int min = 0;int max = x;while (max - min > 1) {int m = (max + min) / 2;if (x / m < m) {max = m;} else {min = m;}}return min;}
}

要点

方案一和方案二是直接求解。
方案三，使用牛顿迭代法，不过实现并不优雅。
方案四，使用二分查找或者折半查找，不过实现并不优雅，使用除法来协助判定当前迭代的值是否满足要求，应该可以规避溢出的问题。
方案五，同样使用了二分查找，相对要优雅一点。

在实现时，需要注意边界值，比如从0，1，2，3，4以及46342、46340。
整型最大值的平方根，取整之后为46340，因此可以作为迭代的上限值。

准备的用例，如下

@Before
public void before() {t = new Solution();
}@Test
public void test001() {assertEquals(2, t.mySqrt(8));
}@Test
public void test002() {assertEquals(2, t.mySqrt(4));
}@Test
public void test003() {assertEquals(1, t.mySqrt(3));
}@Test
public void test004() {assertEquals(1, t.mySqrt(2));
}@Test
public void test005() {assertEquals(2, t.mySqrt(5));
}@Test
public void test006() {assertEquals(46340, t.mySqrt(2147483647));
}