add solution: house robber ii

Author: obzva

house-robber-ii/flynn.go

@@ -0,0 +1,51 @@
+/*
+풀이
+- house robber 문제와 비슷합니다
+  이전 풀이 링크: https://github.com/DaleStudy/leetcode-study/pull/576/files#diff-a98dce0d933d299b3b8e2cc345b95a398c894391b7b86b4b85e3c0aea9d0757f
+- 첫번째 집을 터는 경우와 안 터는 경우를 나누어 계산합니다
+Big O
+- N: 주어진 배열 nums의 길이
+- Time complexity: O(N)
+- Space complexity: O(1)
+*/
+
+import "slices"
+
+func rob(nums []int) int {
+	n := len(nums)
+
+	if n < 4 {
+		return slices.Max(nums)
+	} else if n == 4 {
+		return max(nums[0]+nums[2], nums[1]+nums[3])
+	}
+
+	// rob nums[0] (can't rob nums[1] and nums[n-1])
+	robFirst := nums[0]
+	ppRobbed := nums[2] // initial value = nums[2] because we can't rob nums[1]
+	pRobbed := nums[3]
+	pUnrobbed := nums[2]
+	for i := 4; i < n-1; i++ { // i < n-1 because we can't rob nums[n-1]
+		ppRobbed, pRobbed, pUnrobbed = pRobbed, nums[i]+max(ppRobbed, pUnrobbed), max(pRobbed, pUnrobbed)
+	}
+	robFirst += max(pRobbed, pUnrobbed)
+
+	// skip nums[0]
+	ppRobbed = nums[1]
+	pRobbed = nums[2]
+	pUnrobbed = nums[1]
+	for i := 3; i < n; i++ {
+		ppRobbed, pRobbed, pUnrobbed = pRobbed, nums[i]+max(ppRobbed, pUnrobbed), max(pRobbed, pUnrobbed)
+	}
+	skipFirst := max(pRobbed, pUnrobbed)
+
+	return max(robFirst, skipFirst)
+}
+
+func max(a, b int) int {
+	if a > b {
+		return a
+	} else {
+		return b
+	}
+}