• java中将一个数组按照每份固定大小拆分


    stackoverflow代码

    public static boolean isMatch(byte[] pattern, byte[] input, int pos) {
        for(int i=0; i< pattern.length; i++) {
            if(pattern[i] != input[pos+i]) {
                return false;
            }
        }
        return true;
    }
    
    public static List<byte[]> split(byte[] pattern, byte[] input) {
        List<byte[]> l = new LinkedList<byte[]>();
        int blockStart = 0;
        for(int i=0; i<input.length; i++) {
           if(isMatch(pattern,input,i)) {
              l.add(Arrays.copyOfRange(input, blockStart, i));
              blockStart = i+pattern.length;
              i = blockStart;
           }
        }
        l.add(Arrays.copyOfRange(input, blockStart, input.length ));
        return l;
    }

    引申出来的泛型代码:

        public static <T> boolean isMatch0(T[] pattern, T[] input, int position)
        {
            for(int i=0; i< pattern.length; i++) {
                if(pattern[i] != input[position+i]) {
                    return false;
                }
            }
            return true;
        }
        public static <T> List<T[]> split(T[] pattern, T[] input)
        {
            List<T[]> l = new LinkedList<T[]>();
            int blockStart = 0;
            for(int i=0; i<input.length; i++) {
                if(isMatch0(pattern,input,i)) {
                    l.add(Arrays.copyOfRange(input, blockStart, i));
                    blockStart = i+pattern.length;
                    i = blockStart;
                }
            }
            l.add(Arrays.copyOfRange(input, blockStart, input.length ));
            return l;
        }

    转载自:https://stackoverflow.com/questions/22519346/how-to-split-a-byte-array-around-a-byte-sequence-in-java

        public <T> List<T[]> split(T[] original, int size) {
            List<T[]> result = new ArrayList<T[]>();
            if (original == null) return result;
    
            int curPos = 0;
            while (curPos < original.length) {
                int remaining = original.length - curPos+1;
                if (remaining <= size) {
                    result.add(Arrays.copyOfRange(original,curPos,original.length));
                } else {
                    result.add(Arrays.copyOfRange(original,curPos,curPos+size));
                }
                curPos += size;
            }
    
            return result;
        }
    byte[] data = { 2, 3, 5, 7, 8, 9, 11, 12, 13 };
    
    int blockSize = 3;
    int blockCount = (data.length + blockSize - 1) / blockSize;
    
    byte[] range = null;
    
    for (int i = 1; i < blockCount; i++) {
            int idx = (i - 1) * blockSize;
            range = Arrays.copyOfRange(data, idx, idx + blockSize);
            System.out.println("Chunk " + i + ": " Arrays.toString(range));
    }
    
    // Last chunk
    int end = -1;
    if (data.length % blockSize == 0) {
            end = data.length;
    } else {
            end = data.length % blockSize + blockSize * (blockCount - 1);
    }
            
    range = Arrays.copyOfRange(data, (blockCount - 1) * blockSize, end);
    
    System.out.println("Chunk " + blockCount + ": " Arrays.toString(range));

    https://gist.github.com/lesleh/7724554

    以上代码不能分割byte数组,分割byte数组如下

    byte[] data = { 2, 3, 5, 7, 8, 9, 11, 12, 13 };
    
    int blockSize = 3;
    int blockCount = (data.length + blockSize - 1) / blockSize;
    
    byte[] range = null;
    
    for (int i = 1; i < blockCount; i++) {
            int idx = (i - 1) * blockSize;
            range = Arrays.copyOfRange(data, idx, idx + blockSize);
            System.out.println("Chunk " + i + ": " Arrays.toString(range));
    }
    
    // Last chunk
    int end = -1;
    if (data.length % blockSize == 0) {
            end = data.length;
    } else {
            end = data.length % blockSize + blockSize * (blockCount - 1);
    }
            
    range = Arrays.copyOfRange(data, (blockCount - 1) * blockSize, end);
    
    System.out.println("Chunk " + blockCount + ": " Arrays.toString(range));

    https://gist.github.com/lesleh/7724554

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  • 原文地址:https://www.cnblogs.com/passedbylove/p/12462613.html
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