Implement int sqrt(int x).
Compute and return the square root of x.
我采用的是二分法。每次折中求平方,如果大了就把中值赋给大的,如果小了就把中值赋给小的。
public int mySqrt(int x) {
long start = 1, end = x;
while (start + 1 < end) {
long mid = start + (end - start) / 2;
if (mid * mid <= x) {
start = mid;
} else {
end = mid;
}
}
if (end * end <= x) {
return (int)end;
}
return (int) start;
}