代码随想录二刷day44

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前言

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一、力扣动态规划：完全背包理论基础

//先遍历物品，再遍历背包
private static void testCompletePack(){
    int[] weight = {1, 3, 4};
    int[] value = {15, 20, 30};
    int bagWeight = 4;
    int[] dp = new int[bagWeight + 1];
    for (int i = 0; i < weight.length; i++){ // 遍历物品
        for (int j = weight[i]; j <= bagWeight; j++){ // 遍历背包容量
            dp[j] = Math.max(dp[j], dp[j - weight[i]] + value[i]);
        }
    }
    for (int maxValue : dp){
        System.out.println(maxValue + "   ");
    }
}

//先遍历背包，再遍历物品
private static void testCompletePackAnotherWay(){
    int[] weight = {1, 3, 4};
    int[] value = {15, 20, 30};
    int bagWeight = 4;
    int[] dp = new int[bagWeight + 1];
    for (int i = 1; i <= bagWeight; i++){ // 遍历背包容量
        for (int j = 0; j < weight.length; j++){ // 遍历物品
            if (i - weight[j] >= 0){
                dp[i] = Math.max(dp[i], dp[i - weight[j]] + value[j]);
            }
        }
    }
    for (int maxValue : dp){
        System.out.println(maxValue + "   ");
    }
}

二、力扣518. 零钱兑换 II

class Solution {
    public int change(int amount, int[] coins) {
        int[] dp = new int[amount+1];
        dp[0] = 1;
        for(int i = 0; i < coins.length; i ++){
            for(int j = coins[i]; j <= amount; j ++){
                dp[j] += dp[j-coins[i]];
            }
        }
        return dp[amount] == 0 ? 0 : dp[amount];
    }
}

三、力扣377. 组合总和 Ⅳ

class Solution {
    public int combinationSum4(int[] nums, int target) {
        int[] dp = new int[target+1];
        dp[0] = 1;
        for(int i = 0; i <= target; i ++){
            for(int j = 0; j < nums.length; j ++){
                if(i >= nums[j]){
                    dp[i] += dp[i-nums[j]];
                }
            }
        }
        return dp[dp.length-1];
    }
}