Knapsack Problem
In 0/1 knapsack, each item has a weight and value and may be chosen at most once. The goal is maximum total value without exceeding capacity.
先从这个心智模型开始
In 0/1 knapsack, each item has a weight and value and may be chosen at most once. The goal is maximum total value without exceeding capacity.
它如何工作，不堆砌不必要的术语
A DP state can represent the best value using the first i items with capacity c.
For each item, either skip it or take it if it fits.
把算法读成一连串决策
Define the state precisely.
Write the transition/choice from smaller states.
Set base cases, then decide memoization, tabulation or backtracking.
Read a tiny input by hand and write down the state after every important operation. For Knapsack Problem, the goal is to make the invariant visible rather than to memorize code.
function knapsack(items, capacity) {
const dp = Array(capacity + 1).fill(0);
for (const { weight, value } of items) {
for (let c = capacity; c >= weight; c--) {
dp[c] = Math.max(dp[c], dp[c - weight] + value);
}
}
return dp[capacity];
}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(nW) DP
Do not quote this complexity mechanically.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
- Use DP when subproblems overlap and optimal solutions can be composed from smaller optimal states.
通常哪里会出错
- Writing recurrence before defining state.
- Using DP when state space is larger than the original problem.
专业回答结构
Clarify the variant first—0/1, unbounded or fractional. Their correct algorithms differ substantially.
What problem does Knapsack Problem solve?
In 0/1 knapsack, each item has a weight and value and may be chosen at most once. The goal is maximum total value without exceeding capacity.
关于复杂度应该提到什么？
Its typical cost here is O(nW) DP, 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？
Writing recurrence before defining state.