分析
已知 sqrt(2)约等于 1.414,那么就可以在(1.4, 1.5)区间做二分查找,
如: high=1.5 low=>1.4 mid => (high+low)/2=1.45
判断 1.45*1.45>2 ? high=1.45 : low =>1.45
退出条件:前后两次的差值的绝对值<=0.0000000001, 则可退出
const double EPSINON = 0.0000000001;
double sqrt2( ){
double low = 1.4, high = 1.5;
double mid = (low + high) / 2;
while (high - low > EPSINON){
if (mid*mid > 2){
high = mid;
}
else{
low = mid;
}
mid = (high + low) / 2;
}
return mid;
}