算法面试——动态规划:0-1背包、最长子序列
一、0-1 背包
public int knapsack(int[] weights, int[] values, int capacity) {int n weights.length;int[][] dp new int[n 1][capacity 1];for (int i 1; i < n; i) {for (int w 1; w < capacity; w) {if (weights[i-1] > w) {dp[i][w] dp[i-1][w];} else…
2026/8/10 6:24:03