Array assignment

Catalog

1. Define a length of 10 Integer array nums , Loop input 10 It's an integer . And then lose Enter an integer , Find this integer , Find the output subscript , I didn't find the prompt .

2. Define a length of 10 Integer array nums , Loop input 10 It's an integer .

The maximum value of the output array 、 minimum value .

3. Please find the sum in the array as The two integers of the target value , And output their array index Suppose that each input corresponds to only one answer , You can't reuse the same elements in this array . Example : Given nums = [2, 7, 11, 15], target = 9 because nums[0] + nums[1] = 2 + 7 = 9 So the output 0,1

4. An array {1,3,9,5,6,7,15,4,8} Sort , Then use dichotomy to find the element 6 And output the sorted subscript .

 5. Move zero ： Array {0,1,0,3,12} Various methods of

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1. Define a length of 10 Integer array nums , Loop input 10 It's an integer .
And then lose Enter an integer , Find this integer , Find the output subscript , I didn't find the prompt .

 int t = -1;// Judge whether the subscript has changed the programming idea , because t The future represents the subscript , The subscript will never be -1,
        Scanner input = new Scanner(System.in);
        System.out.println(" Please enter the first 10 A positive integer ");
        int[]nums = new int[10];
        for (int i=0;i<nums.length;i++){
            nums[i] = input.nextInt();
        }
        System.out.println(" Please enter a positive integer ");
        int n = input.nextInt();
        for (int i = 0;i<nums.length;i++){
            if (n==nums[i]){
                t=i;
                System.out.println(" Integers  " + n + " In the array, the subscript is " + t);
                // Nothing here return, Ensure that the same data can be output 
            }
        }
        if (t==-1){// still -1, Indicates that the number of times is not found 
            System.out.println(" There are no integers in the array " + n);
        }
    }
}

2. Define a length of 10 Integer array nums , Loop input 10 It's an integer .

The maximum value of the output array 、 minimum value .

 System.out.println(" Please enter ten integers ：");
        Scanner input=new Scanner(System.in);
        int[]nums = new int[10];
        for (int i=0;i<nums.length;i++){
            nums[i]=input.nextInt();
        }
        int max = nums[0];
        int min = nums[0];
        for (int i =0;i<nums.length;i++){
            if (max<nums[i]){
                max=nums[i];
            }
            if (min>nums[i]){
                min=nums[i];
            }
        }
        System.out.println(" The maximum value of the array is ："+max);
        System.out.println(" The minimum value of the array is ："+min);
    }
}

3. Please find the sum in the array as The two integers of the target value ,
And output their array index Suppose that each input corresponds to only one answer , You can't reuse the same elements in this array .
Example : Given nums = [2, 7, 11, 15], target = 9
because nums[0] + nums[1] = 2 + 7 = 9 So the output 0,1

int[]nums={2,7,11,15};
        int target=9;
        for(int i =0;i<nums.length;i++) {
            for(int j =0;j<nums.length;j++) {
                if(nums[i]+nums[j]==target&&j!=i) {
                    System.out.println(" The subscript of the two integers of the target value is ："+i+" and "+j);
                    return;

4. An array {1,3,9,5,6,7,15,4,8} Sort ,
Then use dichotomy to find the element 6 And output the sorted subscript .

int nums[] = {1, 3, 9, 5, 6, 7, 15, 4, 8};
        int target = 6;
        System.out.println(" Array nums Before ordering ：");
        for (int num : nums) {
            System.out.print(num + " ");
        }
        System.out.println();
        Arrays.sort(nums);// Sort the input array of integers 

        System.out.println(" Array nums After ordering ：");
        for (int num : nums) {
            System.out.print(num + " ");
        }
        System.out.println();

        // The binary search method is used 
        int minIndex = 0;
        int maxIndex = nums.length - 1;
        int midIndex = 0;
        while (true) {
            midIndex = minIndex + ((maxIndex - minIndex) >> 1);
            if (target > nums[midIndex]) {
                // The intermediate data is large 
                minIndex = midIndex + 1;
            } else if (target < nums[midIndex]) {
                // The intermediate data is small 
                maxIndex = midIndex - 1;
            } else {
                // Find target data , Data location ：midIndex
                break;
            }

            if (minIndex > maxIndex) {
                // Target data not found , return -1
                midIndex = -1;
                break;
            }
        }
        System.out.println(" lookup " + target + " The integer is in the... Of the array subscript " + midIndex);
    }
}

 5. Move zero ： Array {0,1,0,3,12} Various methods of

Method 1： Reassign the method at the end  , Stupid method

int[]nums= {0,1,0,3,12};
        int temp=0;// To calculate 0 The number of 
        for(int i=0;i<nums.length;i++){
            if(nums[i]==0){// If nums[i]=0,temp Number plus one 
                temp++;
            }else if(temp!=0){// If nums[i] It's not equal to 0, take nums[i] In front of you temp Exchange , And will nums[i] The assignment is 0
                nums[i-temp]=nums[i];
                nums[i]=0;
            }
        }
        for(int i =0;i<nums.length;i++) {
            System.out.println(nums[i]);
        }
    }
}

Method 2： Double cycle comparison

int temp;
int nums[]={0,1,0,3,12};
for(int i=0;i<nims.length;i++){
  for(int j=0;j<nims.length;j++){
    if(nums[i]==0&&nums[j]!=0){
       temp=nums[i];
       nums[i]=nums[j];
       nums[j]=temp;

Method 3： Reference new array  

int[] nums = {0, 1, 0, 3, 12};
        int index = 0;// Index of the new array subscript 
        int[] nums2 = new int[nums.length];
        for (int i = 0; i < nums.length; i++) {
            if (nums[i] != 0) {
                nums2[index++] = nums[i];// This step is the core , Most difficult to think of 
            }
        }
        for (int i = 0; i < nums2.length; i++) {
            System.out.println(nums2[i]);
        }
    }
}