leetcode66 將數組表示的非負整數加一

題目要求：一個非負整數被表示爲一個數組，數組中每個元素表明該整數的一個位。數組的下標越小，表明的位數越高。如今對該數組作加一運算，請返回結果數組。

/**
 * @author rale
 *
 * Given a non-negative integer represented as a non-empty array of digits, plus one to the integer.
 * You may assume the integer do not contain any leading zero, except the number 0 itself.
 * The digits are stored such that the most significant digit is at the head of the list.
 */
public class PlusOne {

    public int[] plusOne(int[] digits) {
        //此處能夠直接將carry（進位）設置爲1，優化程序
        //carry = 0
        //digits[digits.length-1] += 1 ;
        int carry = 1;
        int temp = 0;
        for(int i=digits.length-1 ; i>=0 ; i--){
            temp = digits[i] + carry;
            digits[i] = temp%10;
            carry = temp/10;
        }
        if(carry>0){
            int[] result = new int[digits.length+1];
            result[0] = 1;
            for(int j = 1 ; j<result.length ; j++){
                result[j] = digits[j-1];
            }
            return result;
        }
        return digits;
    }
}

繼續優化
只有當須要進位的時候，加法才須要繼續下去，不然加法則能夠在當前位中止。
能夠在循環中添加判斷，若carry==0，則提早跳出循環

/**
 * @author rale
 *
 * Given a non-negative integer represented as a non-empty array of digits, plus one to the integer.
 * You may assume the integer do not contain any leading zero, except the number 0 itself.
 * The digits are stored such that the most significant digit is at the head of the list.
 */
public class PlusOne {

    public int[] plusOne(int[] digits) {
        //此處能夠直接將carry（進位）設置爲1，優化程序
        //carry = 0
        //digits[digits.length-1] += 1 ;
        int carry = 1;
        int temp = 0;
        for(int i=digits.length-1 ; i>=0 ; i--){
            temp = digits[i] + carry;
            digits[i] = temp%10;
            carry = temp/10;
            if(carry==0){
                break
            }
        }
        if(carry>0){
            int[] result = new int[digits.length+1];
            result[0] = 1;
            for(int j = 1 ; j<result.length ; j++){
                result[j] = digits[j-1];
            }
            return result;
        }
        return digits;
    }
}

再再再次優化
此處優化最高位進位的狀況
最高位出現進位，當且僅當其餘位都產生進位且爲0
優化後的代碼以下

/**
 * @author rale
 *
 * Given a non-negative integer represented as a non-empty array of digits, plus one to the integer.
 * You may assume the integer do not contain any leading zero, except the number 0 itself.
 * The digits are stored such that the most significant digit is at the head of the list.
 */
public class PlusOne {

    public int[] plusOne(int[] digits) {
        int carry = 1;
        int temp = 0;
        for(int i=digits.length-1 ; i>=0 ; i--){
            temp = digits[i] + carry;
            digits[i] = temp%10;
            carry = temp/10;
        }
        if(carry>0){
            int[] result = new int[digits.length+1];
            result[0] = 1;
//          最高位進位的狀況只有一種，即其它位均進位且爲0，無需再循環一次
//            for(int j = 1 ; j<result.length ; j++){
//                result[j] = digits[j-1];
//            }
            return result;
        }
        return digits;
    }    
}

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