Integer 类中的一些方法

1,计算一个正int数有几位的方法：

亮点：通过维护一个数组提高性能

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????final?static?int?[]?sizeTable?=?{?9,?99,?999,?9999,?99999,?999999,?9999999,

??????????????????????????????????????99999999,?999999999,?Integer.MAX_VALUE?};

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????//?Requires?positive?x

????static?int?stringSize(int?x)?{

????????for?(int?i=0;?;?i++)

????????????if?(x?<=?sizeTable[i])

????????????????return?i+1;

}

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2,?int?值的位数放到对应的数组?buf

亮点：通过位移的方式替代十进制计算的方式，提高性能。

static?void?getChars(int?i,?int?index,?char[]?buf)?{

????????int?q,?r;

????????int?charPos?=?index;

????????char?sign?=?0;

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????????if?(i?<?0)?{?

????????????sign?=?'-';

????????????i?=?-i;

????????}

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????????//?Generate?two?digits?per?iteration

????????while?(i?>=?65536)?{

????????????q?=?i?/?100;

????????//?really:?r?=?i?-?(q?*?100);

????????????r?=?i?-?((q?<<?6)?+?(q?<<?5)?+?(q?<<?2));

????????????i?=?q;

????????????buf?[--charPos]?=?DigitOnes[r];

????????????buf?[--charPos]?=?DigitTens[r];

????????}

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????????//?Fall?thru?to?fast?mode?for?smaller?numbers

????????//?assert(i?<=?65536,?i);

????????for?(;;)?{?

????????????q?=?(i?*?52429)?>>>?(16+3);

????????????r?=?i?-?((q?<<?3)?+?(q?<<?1));??//?r?=?i-(q*10)?...

????????????buf?[--charPos]?=?digits?[r];

????????????i?=?q;

????????????if?(i?==?0)?break;

????????}

????????if?(sign?!=?0)?{

????????????buf?[--charPos]?=?sign;

????????}

}

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