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

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转载自www.cnblogs.com/passedbylove/p/12462613.html