几乎递增的序列java

Question

我正在尝试编写代码来确定是否可以通过仅从该数组中删除一个元素来获得一个严格递增的整数数组。

我的代码适用于 17 种情况中的 16 种，但我想不出一种方法来巧妙地重写我的代码，以便解决数字大于之前的数字以及小于之后的数字的情况这是我写这个for循环的方式。这是我的代码。这不适用于数组：[1, 2, 3, 4, 3, 6]，因为它不会像我的 for 循环当前构造的方式那样将数组中的最后 3 个视为违规者.

boolean almostIncreasingSequence(int[] sequence) {

int offenderPosition = 0;
int[] arrCopy = Arrays.copyOf(sequence, sequence.length);
boolean ordered = true;


//trying to neatly rewrite this for loop 
for(int i= 0; i < sequence.length; i++){
    if(i<sequence.length-1){
        for(int j = i+1; j < sequence.length; j++) {
            if(!(sequence[i] < sequence[j])){
                ordered = false;
                offenderPosition = i;
            }
        }
    }
    if(i == sequence.length-1){
        if(!(sequence[i] > sequence[i-1])){
            ordered = false;
            offenderPosition = i; 
        }
    }

}


if(ordered == false) {
    //remove offender 
    int currentSize = arrCopy.length;
    for(int i = offenderPosition+1;i< currentSize; i++) {
        arrCopy[i-1] = arrCopy[i];
    }
    currentSize--;

    //reassign array
    arrCopy = Arrays.copyOf(arrCopy, currentSize);

    boolean lastChance = true;

    for(int i = 0; i < currentSize-1; i++){
        for(int j = i+1; j < currentSize; j++) {
            if(!(arrCopy[i] < arrCopy[j])){
                lastChance = false;
            }
        }
    }
    return lastChance;
}
else{
    return true;
}

}

Answer 1

我认为这可能有效：

boolean almostIncreasingSequence(int[] a) {
    int count1 = 0 , count2 = 0;
    for(int i = 0 ; i < a.length-1 ; i++){
        if(a[i] >= a[i+1]) count1++;
    }

    for(int i = 0 ; i < a.length-2 ; i++){
        if(a[i] >= a[i+2]) count2++;
    }
     return (count1 <=1) && (count2 <= 1);
}

第一个循环只检查彼此接近的数字。如果第一个索引大于第二个索引，我们将在 count1 中加 1。将 1 添加到 count1 时，表示第一个索引大于第二个索引，该方法应返回 false；第二个 for 循环也将检查 ex。如果第一个索引大于第三个索引。 1, 2 ,1,2 例如，它会将 1 添加到 count2 每个循环执行后，该方法将返回 if 语句返回的布尔值。

Answer 2

你可以把代码分解成几个方法：

// The first method just checks if the input array is sorted
public static boolean isAscending(int[] arr) {

    boolean sorted = true;
    for (int i = 0; i < arr.length - 1; ++i) {
        if (arr[i] >= arr[i + 1]) {
            sorted = false;
            break;
        }
    }
    return sorted;
}

// The second method is the important one.
public static boolean isAlmostAscending(int[] array) {
    int[] tmpArray = new int[array.length - 1];
    // loop through all possible combinations
    for(int i = 0; i < array.length; ++i) {
        copyArray(array, tmpArray, i);
        if(isAscending(tmpArray)) {
            // if the array is sorted after skipping element i, we are done
            return true;
        }
    }
    return false;
}

// helper method to copy array and skip element at skip
private static void copyArray(int[] srcArray, int[] destArray, int skip) {
    for(int i = 0, j = 0; i < destArray.length; ++i, ++j) {
        if(i == skip) {
            ++j;
        }
        destArray[i] = srcArray[j];
    }
}

您可以将所有三种方法合二为一，如下所示：

public static boolean isAlmostAscending(int[] array) {
    int[] tmpArray = new int[array.length - 1];
    // loop through all possible combinations
    for(int index = 0; index < array.length; ++index) {

        // copyArray
        for(int i = 0, j = 0; i < tmpArray.length; ++i, ++j) {
            if(i == index) {
                ++j;
            }
            tmpArray[i] = array[j];
        }

        // check if the current array is sorted
        boolean sorted = true;
        for (int i = 0; i < tmpArray.length - 1; ++i) {
            if (tmpArray[i] >= tmpArray[i + 1]) {
                sorted = false;
                break;
            }
        }
        if(sorted) {
            // if the array is sorted after skipping element i, we are done
            return true;
        }
    }
    return false;
}

Answer 3

解决办法如下：

  boolean almostIncreasingSequence(int[] sequence) {
       int count = checkSeq(sequence);
     if(count == -1) {
         return true;
     }
     int index = count;
     count = checkSeq(removeIndex(sequence,count-1));
    if(count == -1) {
         return true;
     }
     count = checkSeq(removeIndex(sequence,index));
if(count == -1) {
         return true;
     }
return false;
}
int checkSeq(int[] source){
    int index = -1;
     for(int i=1; i<source.length;i++){
           if(source[i] <= source[i-1]){
             index = i;
             break;
         }
     }

    return index;
}

int[] removeIndex(int[] source , int index){
    int[] dist = new int[source.length-1];
    int x = 0;
    for(int i=0; i<source.length;i++){
         if(i==index){
            continue;
        }
        dist[x++] = source[i];
    }
    return dist;
}

Answer 4

boolean almostIncreasingSequence(int[] sequence) {
    int count=0;
    for(int i=0;i<sequence.length-1;i++){
        //if a problem is found increase the counter
        if(sequence[i]>=sequence[i+1]){
            count++;
            /*Lots of conditions, just making sure we don't go out of bounds so
              that we can check if there are any potential duplicates/order issues if 
            a number would be removed
            */
            if(i-1>=0&&i+2<sequence.length&&sequence[i- 
            1]>=sequence[i+1]&&sequence[i]>=sequence[i+2]){
                return false;
            }
            //if we found a problem twice
            if(count>=2){
                return false;
            }
        }
    } return true;
}