LeetCode--66. 加一

没太大难度，仅需注意9+1后的进位情况

public class Q66_Plus_One {
    public static void main(String[] args) {
        System.out.println(Arrays.toString(plusOne(new int[]{1, 2, 3})));
        System.out.println(Arrays.toString(plusOne(new int[]{4, 3, 2, 1})));
        System.out.println(Arrays.toString(plusOne(new int[]{0})));
        System.out.println(Arrays.toString(plusOne(new int[]{1, 9})));
        System.out.println(Arrays.toString(plusOne(new int[]{9, 9})));
    }

    public static int[] plusOne(int[] digits) {
        //进位
        int plus = 1;
        for (int i = digits.length - 1; i >= 0; i--) {
            if (digits[i] == 9 && plus == 1) {
                digits[i] = 0;
            } else {    //只要有一位不进位，循环就结束了，因为剩余的值不会变动了
                digits[i] += 1;
                plus = 0;
                break;
            }
        }
        if (plus == 0) {
            return digits;
        } else {    //{9, 9}+1 -> {1, 0, 0}
            int[] plusDigits = new int[digits.length + 1];
            plusDigits[0] = 1;
            for (int i = 1; i < plusDigits.length; i++) {
                plusDigits[i] = 0;
            }
            return plusDigits;
        }
    }
}

欢迎讨论

标签：digits,加一,--,System,int,new,plusOne,LeetCode,out
来源： https://blog.csdn.net/asd2671193566/article/details/116699282