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Power-n

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	`1`	`+/*`
	`2`	`+ Author: King, wangjingui@outlook.com`
	`3`	`+ Date: Dec 25, 2014`
	`4`	`+ Problem: Pow(x, n)`
	`5`	`+ Difficulty: easy`
	`6`	`+ Source: https://oj.leetcode.com/problems/powx-n/`
	`7`	`+ Notes:`
	`8`	`+ Implement pow(x, n).`
	`9`	`+`
	`10`	`+ Solution: recursion.`
	`11`	`+ */`
	`12`	`+publicclassSolution {`
	`13`	`+publicdoublepow(doublex,intn) {`
	`14`	`+if (x <0)return (n %2 ==0) ?pow(-x,n) : -pow(-x,n);`
	`15`	`+if (x ==0 \|\|x ==1)returnx;`
	`16`	`+if (n <0)return1.0 /pow(x,-n);`
	`17`	`+if (n ==0)return1.0;`
	`18`	`+if (n ==1)returnx;`
	`19`	`+doublehalf =pow(x,n/2);`
	`20`	`+if (n %2 ==0)returnhalf *half;`
	`21`	`+elsereturnx half half;`
	`22`	`+ }`
	`23`	`+}`

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