66. plus one

Title Description

Given a non-negative integer nonempty array of integers represented, on the basis of the number plus one.
Most significant digit stored in the first array, each element of the array stores only a single digit.
You may assume that in addition to the integer 0, the integer does not begin with a zero.

Example:

输入: [1,2,3]
输出: [1,2,4]
解释: 输入数组表示数字 123。

Links: https://leetcode-cn.com/problems/plus-one

Ideas analysis

Reference clever solution to a problem the god of solution: the Java math problem solving
a number with a storage array, then add only two cases after 1

9 is not the last one, last one added directly to the array after the array returns 1.
9 is the last case, it is necessary to determine the cycle, because it's in front of a possibly 9, let 9 to 0, then the former plus 1.
If the highest level 9, you can create a new array, the array length than the original length freshman, the first position 0. To

    /**
     * 将一个用表示数字的数组传入，加上1之后，返回新数组
     * @param digits  传入的数组
     * @return  返回加一之后的新数组
     */
    public static int[] plusOne(int[] digits){
        int len = digits.length;
        for(int i = len-1;i>=0;i--){
            digits[i]++;
            digits[i] = digits[i]%10;
            //末位不是9，直接返回
            if(digits[i]!=0) return digits;
        }
        //最高位为9的情况，创建一个比原数组长1的数组
        int[] newDigits = new int[len+1];
        //首位置1，其余初始化为0
        newDigits[0] = 1;
        return newDigits;
    }

88. merge two ordered arrays

Title Description

Given two ordered arrays of integers and nums2 nums1, incorporated into the nums2 nums1 in an ordered array such as a num1.

Description:

And initializing the number of elements nums1 nums2 of m and n.
You can assume nums1 sufficient space (space equal to or greater than m + n) to save the elements of nums2.

Example:

    输入:
    nums1 =[1,2,3,0,0,0],m =3
    nums2 =[2,5,6],n =3
    输出: [1,2,2,3,5,6]

Links: https://leetcode-cn.com/problems/merge-sorted-array

Ideas analysis

Reference Solution: painting solution algorithm

Since there is enough space nums1 support under two arrays, the processing from back to front merge sort added .
Set point two arrays of digital tail pointers p1 and p2, and p is a pointer to a pointer to merge just the last bit of the array.
Comparing two values from back to front in the array, back into the large nums1 sequentially fill forward.
When no time nums1 number , all copies of the nums2 nums1 the front.

    /**
     * 合并并排序到第一个数组
     *
     * @param nums1 传入的第一个数组
     * @param m     第一个数组的长度
     * @param nums2 传入的第二个数组
     * @param n     第二个数组的长度
     */
    public static void merge(int[] nums1, int m, int[] nums2, int n) {
        int p1 = m - 1;//指向nums1的有数字的最后一位
        int p2 = n - 1;//指向nums2的有数字的最后一位
        int p = m + n - 1;
        //从有数字的最后一位向前，一一对比，两者较大的数存到nums1的p位置，分别向前
        while (p1 >= 0 && p2 >= 0) {
            nums1[p--] = nums1[p1] > nums2[p2] ? nums1[p1--] : nums2[p2--];
        }
        //由于是存到nums1上的，所以保证最后p1无元素可指的时候，把nums2上的元素按顺序拷贝到nums1
        System.arraycopy(nums2, 0, nums1, 0, p2 + 1);
    }

167. The two numbers and enter an ordered array Ⅱ-

Title Description

Has been given a follow ascending ordered array to find the number of two numbers such that their sum is equal to the target sum.
Function should return the two index values and index1 index2, which must be less than index1 index2.

Description:

Return values of the index (index1, index2 and) is not zero .
You can assume that each input corresponding to only the only answer, but you can not reuse the same elements.
Example:

输入: numbers = [2, 7, 11, 15], target = 9
输出: [1,2]
解释: 2 与 7 之和等于目标数 9 。因此 index1 = 1, index2 = 2 。

Links: https://leetcode-cn.com/problems/two-sum-ii-input-array-is-sorted

Ideas analysis

Reference Solution: Dual Pointer solving two numbers

Note that the array is sorted, you can use it to solve.
Solving double pointer , a pointer to the front, a pointer to the rear.
If two pointer corresponds exactly and target, two return pointer index corresponding to +1.
If greater than target, large numbers pointer will move forward . Conversely, the decimal pointer moves rearwardly.

    /**
     * 得到传入数组中两个元素和刚好是目标数字的位置，用数组返回
     * @param numbers   传入的目标数组
     * @param target    传入的目标数字
     * @return   返回两元素和为目标数字的对应位置
     */
    public static int[] twoSum(int[] numbers,int target){
        int i = 0;
        int j = numbers.length-1;
        if(numbers == null) return null;
        while(i<j){
            int sum = numbers[i]+numbers[j];
            if(sum==target) return new int[]{i+1,j+1};
            else if(sum>target){
                j--;
            }else{
                i++;
            }
        }
        return null;
    }