01背包问题 二维 
dp[i][j] 表示在物品i时，背包在j容量下的最大价值，递推公式为

dp[i][j] = Math.max(dp[i-1][j] , dp[i-1][j-weight[i]] + value[i]);

第一个时不放物品i，其价值等于在物品i-1时背包在j容量下的最大价值，第二个是放物品i，表示是在i-1物品时，背包在j-weight[i]时的价值加上i物品的价值。

public class BagProblem {public static void main(String[] args) {int[] weight = {1,3,4};int[] value = {15,20,30};int bagSize = 4;testWeightBagProblem(weight,value,bagSize);}public static void testWeightBagProblem(int[] weight, int[] value, int bagSize){int goods = weight.length;int[][] dp = new int[goods][bagSize + 1];for (int j = weight[0]; j <= bagSize; j++) {dp[0][j] = value[0];}for (int i = 1; i < weight.length; i++) {for (int j = 1; j <= bagSize; j++) {if (j < weight[i]) {dp[i][j] = dp[i-1][j];} else {dp[i][j] = Math.max(dp[i-1][j] , dp[i-1][j-weight[i]] + value[i]);}}}for (int i = 0; i < goods; i++) {for (int j = 0; j <= bagSize; j++) {System.out.print(dp[i][j] + "\t");}System.out.println("\n");}}
}

    public static void main(String[] args) {int[] weight = {1, 3, 4};int[] value = {15, 20, 30};int bagWight = 4;testWeightBagProblem(weight, value, bagWight);}public static void testWeightBagProblem(int[] weight, int[] value, int bagWeight){int wLen = weight.length;//定义dp数组：dp[j]表示背包容量为j时，能获得的最大价值int[] dp = new int[bagWeight + 1];//遍历顺序：先遍历物品，再遍历背包容量for (int i = 0; i < wLen; i++){for (int j = bagWeight; j >= weight[i]; j--){dp[j] = Math.max(dp[j], dp[j - weight[i]] + value[i]);}}//打印dp数组for (int j = 0; j <= bagWeight; j++){System.out.print(dp[j] + " ");}}

class Solution {public boolean canPartition(int[] nums) {if(nums == null || nums.length == 0) return false;int n = nums.length;int sum = 0;for(int num : nums) {sum += num;}//总和为奇数，不能平分if(sum % 2 != 0) return false;int target = sum / 2;int[] dp = new int[target + 1];for(int i = 0; i < n; i++) {for(int j = target; j >= nums[i]; j--) {//物品 i 的重量是 nums[i]，其价值也是 nums[i]dp[j] = Math.max(dp[j], dp[j - nums[i]] + nums[i]);}//剪枝一下，每一次完成內層的for-loop，立即檢查是否dp[target] == target，優化時間複雜度（26ms -> 20ms）if(dp[target] == target)return true;}return dp[target] == target;}
}