From 0ae922af453d7edba2087f4b10339628665a54f5 Mon Sep 17 00:00:00 2001 From: yanglbme Date: Wed, 12 Jun 2024 19:18:13 +0800 Subject: [PATCH] feat: update solutions to lc problem: No.2786 No.2786.Visit Array Positions to Maximize Score --- .../README.md | 31 +++++++++++++------ .../README_EN.md | 31 +++++++++++++------ .../Solution.cpp | 3 +- .../Solution.go | 2 +- .../Solution.java | 3 +- .../Solution.py | 2 +- .../Solution.ts | 8 ++--- 7 files changed, 54 insertions(+), 26 deletions(-) diff --git a/solution/2700-2799/2786.Visit Array Positions to Maximize Score/README.md b/solution/2700-2799/2786.Visit Array Positions to Maximize Score/README.md index 746d86d842f7c..5a5ed886954fc 100644 --- a/solution/2700-2799/2786.Visit Array Positions to Maximize Score/README.md +++ b/solution/2700-2799/2786.Visit Array Positions to Maximize Score/README.md @@ -67,7 +67,18 @@ tags: -### 方法一 +### 方法一：动态规划 + +根据题目描述，我们可以得到以下结论： + +1. 从位置 $i$ 移动到位置 $j$ 时，如果 $nums[i]$ 和 $nums[j]$ 的奇偶性不同，那么会损失 $x$ 分； +2. 从位置 $i$ 移动到位置 $j$ 时，如果 $nums[i]$ 和 $nums[j]$ 的奇偶性相同，那么不会损失分数。 + +因此，我们可以用一个长度为 $2$ 的数组 $f$ 来表示当前位置的奇偶性为 $0$ 和 $1$ 时的最大得分。初始时 $f$ 的值为 $-\infty$，然后我们再初始化 $f[nums[0] \& 1] = nums[0]$，表示初始位置的得分。 + +接下来，我们从位置 $1$ 开始遍历数组 $nums$，对于每个位置 $i$ 对应的值 $v$，我们更新 $f[v \& 1]$ 的值为 $f[v \& 1]$ 和 $f[v \& 1 \oplus 1] - x$ 的较大值再加上 $v$，即 $f[v \& 1] = \max(f[v \& 1], f[v \& 1 \oplus 1] - x) + v$。 + +答案为 $f[0]$ 和 $f[1]$ 中的较大值。 @@ -79,7 +90,7 @@ class Solution: f = [-inf] * 2 f[nums[0] & 1] = nums[0] for v in nums[1:]: - f[v & 1] = max(f[v & 1] + v, f[v & 1 ^ 1] + v - x) + f[v & 1] = max(f[v & 1], f[v & 1 ^ 1] - x) + v return max(f) ``` @@ -92,7 +103,8 @@ class Solution { Arrays.fill(f, -(1L << 60)); f[nums[0] & 1] = nums[0]; for (int i = 1; i < nums.length; ++i) { - f[nums[i] & 1] = Math.max(f[nums[i] & 1] + nums[i], f[nums[i] & 1 ^ 1] + nums[i] - x); + int v = nums[i]; + f[v & 1] = Math.max(f[v & 1], f[v & 1 ^ 1] - x) + v; } return Math.max(f[0], f[1]); } @@ -110,7 +122,8 @@ public: f[nums[0] & 1] = nums[0]; int n = nums.size(); for (int i = 1; i < n; ++i) { - f[nums[i] & 1] = max(f[nums[i] & 1] + nums[i], f[nums[i] & 1 ^ 1] + nums[i] - x); + int v = nums[i]; + f[v & 1] = max(f[v & 1], f[v & 1 ^ 1] - x) + v; } return max(f[0], f[1]); } @@ -125,7 +138,7 @@ func maxScore(nums []int, x int) int64 { f := [2]int{-inf, -inf} f[nums[0]&1] = nums[0] for _, v := range nums[1:] { - f[v&1] = max(f[v&1]+v, f[v&1^1]+v-x) + f[v&1] = max(f[v&1], f[v&1^1]-x) + v } return int64(max(f[0], f[1])) } @@ -135,13 +148,13 @@ func maxScore(nums []int, x int) int64 { ```ts function maxScore(nums: number[], x: number): number { - const inf = 1 << 30; - const f: number[] = Array(2).fill(-inf); + const f: number[] = Array(2).fill(-Infinity); f[nums[0] & 1] = nums[0]; for (let i = 1; i < nums.length; ++i) { - f[nums[i] & 1] = Math.max(f[nums[i] & 1] + nums[i], f[(nums[i] & 1) ^ 1] + nums[i] - x); + const v = nums[i]; + f[v & 1] = Math.max(f[v & 1], f[(v & 1) ^ 1] - x) + v; } - return Math.max(f[0], f[1]); + return Math.max(...f); } ``` diff --git a/solution/2700-2799/2786.Visit Array Positions to Maximize Score/README_EN.md b/solution/2700-2799/2786.Visit Array Positions to Maximize Score/README_EN.md index b38632b020df1..c5ea471975abf 100644 --- a/solution/2700-2799/2786.Visit Array Positions to Maximize Score/README_EN.md +++ b/solution/2700-2799/2786.Visit Array Positions to Maximize Score/README_EN.md @@ -67,7 +67,18 @@ The total score is: 2 + 4 + 6 + 8 = 20. -### Solution 1 +### Solution 1: Dynamic Programming + +Based on the problem description, we can draw the following conclusions: + +1. Moving from position $i$ to position $j$, if $nums[i]$ and $nums[j]$ have different parities, then $x$ points will be lost; +2. Moving from position $i$ to position $j$, if $nums[i]$ and $nums[j]$ have the same parity, then no points will be lost. + +Therefore, we can use an array $f$ of length $2$ to represent the maximum score when the current position's parity is $0$ and $1$. Initially, the values of $f$ are $-\infty$, and then we initialize $f[nums[0] \& 1] = nums[0]$, indicating the score at the initial position. + +Next, we start traversing the array $nums$ from position $1$. For each position $i$ corresponding to the value $v$, we update the value of $f[v \& 1]$ to be the larger value between $f[v \& 1]$ and $f[v \& 1 \oplus 1] - x$ plus $v$, i.e., $f[v \& 1] = \max(f[v \& 1], f[v \& 1 \oplus 1] - x) + v$. + +The answer is the larger value between $f[0]$ and $f[1]$. @@ -79,7 +90,7 @@ class Solution: f = [-inf] * 2 f[nums[0] & 1] = nums[0] for v in nums[1:]: - f[v & 1] = max(f[v & 1] + v, f[v & 1 ^ 1] + v - x) + f[v & 1] = max(f[v & 1], f[v & 1 ^ 1] - x) + v return max(f) ``` @@ -92,7 +103,8 @@ class Solution { Arrays.fill(f, -(1L << 60)); f[nums[0] & 1] = nums[0]; for (int i = 1; i < nums.length; ++i) { - f[nums[i] & 1] = Math.max(f[nums[i] & 1] + nums[i], f[nums[i] & 1 ^ 1] + nums[i] - x); + int v = nums[i]; + f[v & 1] = Math.max(f[v & 1], f[v & 1 ^ 1] - x) + v; } return Math.max(f[0], f[1]); } @@ -110,7 +122,8 @@ public: f[nums[0] & 1] = nums[0]; int n = nums.size(); for (int i = 1; i < n; ++i) { - f[nums[i] & 1] = max(f[nums[i] & 1] + nums[i], f[nums[i] & 1 ^ 1] + nums[i] - x); + int v = nums[i]; + f[v & 1] = max(f[v & 1], f[v & 1 ^ 1] - x) + v; } return max(f[0], f[1]); } @@ -125,7 +138,7 @@ func maxScore(nums []int, x int) int64 { f := [2]int{-inf, -inf} f[nums[0]&1] = nums[0] for _, v := range nums[1:] { - f[v&1] = max(f[v&1]+v, f[v&1^1]+v-x) + f[v&1] = max(f[v&1], f[v&1^1]-x) + v } return int64(max(f[0], f[1])) } @@ -135,13 +148,13 @@ func maxScore(nums []int, x int) int64 { ```ts function maxScore(nums: number[], x: number): number { - const inf = 1 << 30; - const f: number[] = Array(2).fill(-inf); + const f: number[] = Array(2).fill(-Infinity); f[nums[0] & 1] = nums[0]; for (let i = 1; i < nums.length; ++i) { - f[nums[i] & 1] = Math.max(f[nums[i] & 1] + nums[i], f[(nums[i] & 1) ^ 1] + nums[i] - x); + const v = nums[i]; + f[v & 1] = Math.max(f[v & 1], f[(v & 1) ^ 1] - x) + v; } - return Math.max(f[0], f[1]); + return Math.max(...f); } ``` diff --git a/solution/2700-2799/2786.Visit Array Positions to Maximize Score/Solution.cpp b/solution/2700-2799/2786.Visit Array Positions to Maximize Score/Solution.cpp index f46acc1a46c0e..a6c83e43b8c28 100644 --- a/solution/2700-2799/2786.Visit Array Positions to Maximize Score/Solution.cpp +++ b/solution/2700-2799/2786.Visit Array Positions to Maximize Score/Solution.cpp @@ -6,7 +6,8 @@ class Solution { f[nums[0] & 1] = nums[0]; int n = nums.size(); for (int i = 1; i < n; ++i) { - f[nums[i] & 1] = max(f[nums[i] & 1] + nums[i], f[nums[i] & 1 ^ 1] + nums[i] - x); + int v = nums[i]; + f[v & 1] = max(f[v & 1], f[v & 1 ^ 1] - x) + v; } return max(f[0], f[1]); } diff --git a/solution/2700-2799/2786.Visit Array Positions to Maximize Score/Solution.go b/solution/2700-2799/2786.Visit Array Positions to Maximize Score/Solution.go index 3e0f6b6af1a58..9a0ed8ee66ce1 100644 --- a/solution/2700-2799/2786.Visit Array Positions to Maximize Score/Solution.go +++ b/solution/2700-2799/2786.Visit Array Positions to Maximize Score/Solution.go @@ -3,7 +3,7 @@ func maxScore(nums []int, x int) int64 { f := [2]int{-inf, -inf} f[nums[0]&1] = nums[0] for _, v := range nums[1:] { - f[v&1] = max(f[v&1]+v, f[v&1^1]+v-x) + f[v&1] = max(f[v&1], f[v&1^1]-x) + v } return int64(max(f[0], f[1])) } \ No newline at end of file diff --git a/solution/2700-2799/2786.Visit Array Positions to Maximize Score/Solution.java b/solution/2700-2799/2786.Visit Array Positions to Maximize Score/Solution.java index 8cc079c29be30..95523eb596781 100644 --- a/solution/2700-2799/2786.Visit Array Positions to Maximize Score/Solution.java +++ b/solution/2700-2799/2786.Visit Array Positions to Maximize Score/Solution.java @@ -4,7 +4,8 @@ public long maxScore(int[] nums, int x) { Arrays.fill(f, -(1L << 60)); f[nums[0] & 1] = nums[0]; for (int i = 1; i < nums.length; ++i) { - f[nums[i] & 1] = Math.max(f[nums[i] & 1] + nums[i], f[nums[i] & 1 ^ 1] + nums[i] - x); + int v = nums[i]; + f[v & 1] = Math.max(f[v & 1], f[v & 1 ^ 1] - x) + v; } return Math.max(f[0], f[1]); } diff --git a/solution/2700-2799/2786.Visit Array Positions to Maximize Score/Solution.py b/solution/2700-2799/2786.Visit Array Positions to Maximize Score/Solution.py index 71aff1444cc22..f6aed0dcbcc31 100644 --- a/solution/2700-2799/2786.Visit Array Positions to Maximize Score/Solution.py +++ b/solution/2700-2799/2786.Visit Array Positions to Maximize Score/Solution.py @@ -3,5 +3,5 @@ def maxScore(self, nums: List[int], x: int) -> int: f = [-inf] * 2 f[nums[0] & 1] = nums[0] for v in nums[1:]: - f[v & 1] = max(f[v & 1] + v, f[v & 1 ^ 1] + v - x) + f[v & 1] = max(f[v & 1], f[v & 1 ^ 1] - x) + v return max(f) diff --git a/solution/2700-2799/2786.Visit Array Positions to Maximize Score/Solution.ts b/solution/2700-2799/2786.Visit Array Positions to Maximize Score/Solution.ts index 45434278455c0..ed8f80850f858 100644 --- a/solution/2700-2799/2786.Visit Array Positions to Maximize Score/Solution.ts +++ b/solution/2700-2799/2786.Visit Array Positions to Maximize Score/Solution.ts @@ -1,9 +1,9 @@ function maxScore(nums: number[], x: number): number { - const inf = 1 << 30; - const f: number[] = Array(2).fill(-inf); + const f: number[] = Array(2).fill(-Infinity); f[nums[0] & 1] = nums[0]; for (let i = 1; i < nums.length; ++i) { - f[nums[i] & 1] = Math.max(f[nums[i] & 1] + nums[i], f[(nums[i] & 1) ^ 1] + nums[i] - x); + const v = nums[i]; + f[v & 1] = Math.max(f[v & 1], f[(v & 1) ^ 1] - x) + v; } - return Math.max(f[0], f[1]); + return Math.max(...f); }