Finish writing math! as soon as possible

(1) Arranging Coins

Idea one: The idea is about two times equation, get arithmetic and the formula is sum = (x + 1) * X/2

So for this question, if we know and, then we can know x = ( -1 + sqrt (8 * n + 1))/2 down rounding.

The code is as follows:

1  Public class Solution {2      Public int arrangecoins (int  n) {3         return (int) (-1 + MATH.SQRT (1 + 8 * (long) n))/2); 4     }5 }

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Problem-solving idea two: Use the given n to keep subtracting ... until n is less than the number to subtract.

The code is as follows:

1  Public classSolution {2      Public intArrangecoins (intN) {3         intresult = 0;4         intAdd = 1;5          while(N >=add) {6N-=add;7add++;8result++;9         }Ten         returnresult; One     } A}

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(2) Factorial Trailing Zeroes

Problem Solving Ideas:

Because all trailing zero comes from factor 5 * 2.

But sometimes a number can have several 5 factors, for example, 25 has two 5 factors, and 125 has three 5 factors. In n! Operation, factor 2 is always sufficient. So we only calculate the 5 factors in all the numbers from 1 to N.

The simplest way to calculate the number of 5 is SUM (n/5^1, n/5^2, n/5^3 ...)

Code One:

1  Public class Solution {2      Public int trailingzeroes (int  n) {3        return n = = 0? 0:n/5 + Trailingzer OES (N/5); 4     }5 }

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Code two:

1  Public classSolution {2      Public intTrailingzeroes (intN) {3         if(N < 1) {4             return0;5         }6         intNumber = 0;7          while(N/5! = 0) {8N/= 5;9Number + =N;Ten         } One         returnNumber ; A     } -}

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(3) Palindrome number

Problem-solving ideas: To determine whether a number is a palindrome number, can not be like a character string as a character comparison!!! So the number reversal if the same as the original value is a palindrome number.

The code is as follows:

1  Public classSolution {2      Public BooleanIspalindrome (intx) {3         if(X < 0) {4             return false;5         }6         returnx = =reverse (x);7     }8      Public intReverseintx) {9         intRST = 0;Ten          while(x! = 0) { OneRST = rst * ten + x% 10; AX/= 10; -         } -         returnrst; the     } -}

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(4) Rectangle Area

Problem Solving Ideas:

Find the coordinate representation of the repeating part of the rectangle, two large rectangular areas added minus the overlapping portions of the area.

The code is as follows:

1  Public classSolution {2      Public intComputearea (intAintBintCintDintEintFintGintH) {3         intleft = Math.max (a,e), right =Math.max (Math.min (c,g), left);4         intBottom = Math.max (b,f), top =Math.max (Math.min (d,h), bottom);5         return(c-a) * (d-b)-(right-left) * (Top-bottom) + (G-E) * (H-F); 6     }7}

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(5) Reverse Integer

The idea of solving problems is simple and clear.

The code is as follows:

1  Public classSolution {2      Public intReverseintx) {3         intresult = 0;4          while(x! = 0) {5             inttail = x 10;6             intNewresult = result * 10 +tail;7             if((newresult-tail)/10! =result) { 8                 return0; 9}//Determine if overflowTenresult =Newresult; OneX/= 10; A         } -         returnresult; -     } the}

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Finish writing math! as soon as possible