return total
}
-// Generate elements in the chain using an iterative algorithm.
-func (h *huffmanEncoder) generateChains(top *levelInfo, list []literalNode) {
- n := len(list)
- list = list[0 : n+1]
- list[n] = maxNode()
-
- l := top
- for {
- if l.nextPairFreq == math.MaxInt32 && l.nextCharFreq == math.MaxInt32 {
- // We've run out of both leafs and pairs.
- // End all calculations for this level.
- // To m sure we never come back to this level or any lower level,
- // set nextPairFreq impossibly large.
- l.lastChain = nil
- l.needed = 0
- l = l.up
- l.nextPairFreq = math.MaxInt32
- continue
- }
-
- prevFreq := l.lastChain.freq
- if l.nextCharFreq < l.nextPairFreq {
- // The next item on this row is a leaf node.
- n := l.lastChain.leafCount + 1
- l.lastChain = &chain{l.nextCharFreq, n, l.lastChain.up}
- l.nextCharFreq = list[n].freq
- } else {
- // The next item on this row is a pair from the previous row.
- // nextPairFreq isn't valid until we generate two
- // more values in the level below
- l.lastChain = &chain{l.nextPairFreq, l.lastChain.leafCount, l.down.lastChain}
- l.down.needed = 2
- }
-
- if l.needed--; l.needed == 0 {
- // We've done everything we need to do for this level.
- // Continue calculating one level up. Fill in nextPairFreq
- // of that level with the sum of the two nodes we've just calculated on
- // this level.
- up := l.up
- if up == nil {
- // All done!
- return
- }
- up.nextPairFreq = prevFreq + l.lastChain.freq
- l = up
- } else {
- // If we stole from below, move down temporarily to replenish it.
- for l.down.needed > 0 {
- l = l.down
- }
- }
- }
-}
-
// Return the number of literals assigned to each bit size in the Huffman encoding
//
// This method is only called when list.length >= 3