• Java实现LeetCode_0007_ReverseInteger


    package javaLeetCode_primary;
    
    import java.util.Scanner;
    /**
     * Given a 32-bit signed integer, reverse digits of an integer. <b>Example 1:
     * <li>Input: 123
     * <li>Output: 321 Example 2:
     * <li>Input: -123
     * <li>Output: -32  Example 3:
     * <li>Input: 120
     * <li>Output: 21
     */
    
    /*
     * Test data: 
     *1534236469->0
     *999999991->199999999
     *2147483647->0
     *2147483642 ->0
     *1000000000->1
     *-2147483648->0
     *-214748364->-463847412
     *901000-> 109
     */
    public class ReverseInteger_7 {
    	public static void main(String[] args) {
    
    		@SuppressWarnings("resource")
    		Scanner input = new Scanner(System.in);
    		System.out.println("Please input a integer:");
    		int x = input.nextInt();
    		System.out.println(reverse_2(x));
    	}// end main
    
    	/**
    	 * Use simple  methods to solve problems.
    	 */
    	public static int reverse_1(int x) {
    
    		int[] arr = new int[10];// Stores the number of bits of a parameter
    		int countDigit = 0;// Calculate the length of the number
    		int cf = 0;// Leave the parameter state
    		int i = 0;
    		// The element that initializes array is 0.
    		for (i = 0; i < arr.length; i++) {
    			arr[i] = 0;
    		} // end for
    		
    		if (x < 0) {
    			cf = 1;
    			x *= -1;
    		} // end if
    
    		//Get the the digit of parameter and reverse/invert  it.
    		i = 0;
    		while (x > 0) {
    			countDigit++;
    			arr[i] = x % 10;
    			x /= 10;
    			i++;
    		} // end while
    
    		// Print the integer
    		i = 0;
    		long  y = 0;
    		for (; i < countDigit; i++) {
    			y = y * 10 + arr[i];
    		} // end for
    
    		//Judge whether  the integer is valid.
    		if(y>2147483647) {
    			x = 0;
    		}else {
    			x = (int)y;
    		}//end if
    		
    		if (cf == 1) {
    			x *= -1;
    		} // end if
    
    		return x;
    
    	}// end reverse()
    	
    	/**
    	 * Answer online
    	 * */
    	public static int reverse_2(int x) {
            int rev = 0;
            while (x != 0) {
                int pop = x % 10;
                x /= 10;
                if (rev > Integer.MAX_VALUE/10 || (rev == Integer.MAX_VALUE / 10 && pop > 7)) return 0;
                if (rev < Integer.MIN_VALUE/10 || (rev == Integer.MIN_VALUE / 10 && pop < -8)) return 0;
                rev = rev * 10 + pop;
            }
            return rev;
        }//end reverse()
    }
    
    
    
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  • 原文地址:https://www.cnblogs.com/a1439775520/p/12947027.html
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