Counting Sort
Counting sort avoids comparisons. It counts how many times each key appears, then reconstructs the output in key order.
先从这个心智模型开始
Counting sort avoids comparisons. It counts how many times each key appears, then reconstructs the output in key order.
它如何工作，不堆砌不必要的术语
It is powerful when keys are integers in a reasonably small range.
Its complexity depends on n plus the key range k, so a huge sparse range can make it wasteful.
把算法读成一连串决策
Understand what one pass/partition/merge guarantees.
Track which region is already finalized.
Reason about comparisons, swaps/moves and extra memory.
Read a tiny input by hand and write down the state after every important operation. For Counting Sort, the goal is to make the invariant visible rather than to memorize code.
function countingSort(a, max = Math.max(...a)) {
const count = Array(max + 1).fill(0);
for (const x of a) count[x]++;
const out = [];
for (let value = 0; value < count.length; value++) {
for (let c = 0; c < count[value]; c++) out.push(value);
}
return out;
}不要死记语法。跟踪每一行中发生变化的状态，并把它与上面描述的不变量联系起来。
O(n + k)
Do not quote this complexity mechanically. Ask which operation or input dimension it describes, then derive the cost from the loops, recursion depth, data-structure operations and any preprocessing assumptions.
适用场景
- Choose based on input size, order, key range, stability and memory limits.
通常哪里会出错
- Comparing only average Big-O.
- Forgetting stability, in-place behavior, recursion depth or key-domain assumptions.
专业回答结构
Explain why it can beat the comparison-sorting lower bound: it uses assumptions about the key domain rather than only pairwise comparisons.
What problem does Counting Sort solve?
Counting sort avoids comparisons. It counts how many times each key appears, then reconstructs the output in key order.
关于复杂度应该提到什么？
Its typical cost here is O(n + k), but a complete answer should connect that cost to the representation, input constraints and the dominant operations.
最常见的错误是什么？
Comparing only average Big-O.