House Robber的解法

题目链接：https://leetcode.com/problems/house-robber/
答案自然是动态规划(Dynamic programming，简称DP)。
代码

func rob1(nums []int) int {
	lth := len(nums)
	if lth == 0 {
		return 0
	}
	dp := make([]int, lth+1)
	dp[1] = nums[0]
	for i := 2; i < lth+1; i++ {
		dp[i] = int(math.Max(float64(nums[i-1] + dp[i-2]), float64(dp[i-1])))
	}
	return dp[lth]
}

变种题House Robber II
代码

func rob2(nums []int) int {
	lth := len(nums)
	if lth == 0 {
		return 0
	} else if lth == 1 {
		return nums[0]
	}
	return int(math.Max(float64(rob1(nums[:lth-1])), float64(rob1(nums[1:lth]))))
}

第一次做的时候，我并不是这么写的。
受到《算法导论》中矩阵链乘法的影响，我采用的是二维视角。
rob1的代码如下：

func rob1(nums []int) int {
	lth := len(nums)
	if lth == 0 {
		return 0
	} else if lth == 1 {
		return nums[0]
	} else if lth == 2 {
		return int(math.Max(float64(nums[0]), float64(nums[1])))
	}
	dp := make([][]int,lth+1)
	for i := 0; i <= lth; i++ {
		dp[i] = make([]int,lth+1)
	}
	// 长度为1的
	for i := 1; i <= lth; i++ {
		dp[i][i] = nums[i-1]
	}
	// 长度为2的
	for i := 1; i < lth; i++ {
		dp[i][i+1] = int(math.Max(float64(nums[i-1]),float64(nums[i])))
	}
	// 长度>=3的
	for l := 3; l <= lth;l++ {
		for i := 1; i <= lth+1 - l; i++ {
			j := i+ l -1
			dp[i][j] = int(math.Max(float64(dp[i][j-1]), float64(dp[i][j-2] + nums[j-1])))
		}
	}
	return dp[1][lth]
}

rob2相比rob1，只有在长度为n的情况下，才需要特殊处理下

func rob2(nums []int) int {
	lth := len(nums)
	if lth == 0 {
		return 0
	} else if lth == 1 {
		return nums[0]
	} else if lth == 2 {
		return int(math.Max(float64(nums[0]), float64(nums[1])))
	}
	dp := make([][]int,lth+1)
	for i := 0; i <= lth; i++ {
		dp[i] = make([]int,lth+1)
	}
	// 长度为1的
	for i := 1; i <= lth; i++ {
		dp[i][i] = nums[i-1]
	}
	// 长度为2的
	for i := 1; i < lth; i++ {
		dp[i][i+1] = int(math.Max(float64(nums[i-1]),float64(nums[i])))
	}
	// 长度>=3的
	for l := 3; l < lth;l++ {
		for i := 1; i <= lth+1 - l; i++ {
			j := i+ l -1
			dp[i][j] = int(math.Max(float64(dp[i][j-1]), float64(dp[i][j-2] + nums[j-1])))
		}
	}
	return int(math.Max(float64(dp[1][lth-1]),float64(dp[2][lth-2] + nums[lth-1])))
}

标签：return,nums,int,House,float64,解法,lth,dp,Robber
来源： https://blog.csdn.net/banshichiqinglangzi/article/details/122832259