|
| 1 | +# 安装栅栏 |
| 2 | + |
| 3 | +> 难度：困难 |
| 4 | +> |
| 5 | +> https://leetcode-cn.com/problems/erect-the-fence/ |
| 6 | +
|
| 7 | +## 题目 |
| 8 | + |
| 9 | +在一个二维的花园中，有一些用 `(x, y)` 坐标表示的树。由于安装费用十分昂贵，你的任务是先用**最短**的绳子围起所有的树。只有当所有的树都被绳子包围时，花园才能围好栅栏。你需要找到正好位于栅栏边界上的树的坐标。 |
| 10 | + |
| 11 | + |
| 12 | +示例 1: |
| 13 | + |
| 14 | +``` |
| 15 | +输入: [[1,1],[2,2],[2,0],[2,4],[3,3],[4,2]] |
| 16 | +输出: [[1,1],[2,0],[4,2],[3,3],[2,4]] |
| 17 | +解释: |
| 18 | +``` |
| 19 | + |
| 20 | + |
| 21 | +示例 2: |
| 22 | + |
| 23 | +``` |
| 24 | +输入: [[1,2],[2,2],[4,2]] |
| 25 | +输出: [[1,2],[2,2],[4,2]] |
| 26 | +解释: |
| 27 | +``` |
| 28 | + |
| 29 | +``` |
| 30 | +即使树都在一条直线上，你也需要先用绳子包围它们。 |
| 31 | +``` |
| 32 | + |
| 33 | +## 解题 |
| 34 | + |
| 35 | +### Jarvis 算法 |
| 36 | + |
| 37 | +```ts |
| 38 | +/** |
| 39 | + * Jarvis 算法 |
| 40 | + * @desc 时间复杂度 O(N²) 空间复杂度 O(N) |
| 41 | + * @param trees |
| 42 | + * @returns |
| 43 | + */ |
| 44 | +export function outerTrees(trees: number[][]): number[][] { |
| 45 | + const len = trees.length |
| 46 | + if (len < 4) |
| 47 | + return trees |
| 48 | + |
| 49 | + let leftMost = 0 |
| 50 | + for (let i = 0; i < len; i++) { |
| 51 | + if (trees[i][0] < trees[leftMost][0]) |
| 52 | + leftMost = i |
| 53 | + } |
| 54 | + |
| 55 | + const res: number[][] = [] |
| 56 | + const cross = (p: number[], q: number[], r: number[]) => (q[0] - p[0]) * (r[1] - q[1]) - (q[1] - p[1]) * (r[0] - q[0]) |
| 57 | + |
| 58 | + const visit = new Array(len).fill(false) |
| 59 | + let p = leftMost |
| 60 | + do { |
| 61 | + let q = (p + 1) % len |
| 62 | + for (let r = 0; r < len; r++) { |
| 63 | + // 如果 r 在 pq 右侧，则 q = r |
| 64 | + if (cross(trees[p], trees[q], trees[r]) < 0) |
| 65 | + q = r |
| 66 | + } |
| 67 | + |
| 68 | + // 是否存在点 i，使得 p、q、i 在同一条直线上 |
| 69 | + for (let i = 0; i < len; i++) { |
| 70 | + if (visit[i] || i === p || i === q) |
| 71 | + continue |
| 72 | + |
| 73 | + if (cross(trees[p], trees[q], trees[i]) === 0) { |
| 74 | + res.push(trees[i]) |
| 75 | + visit[i] = true |
| 76 | + } |
| 77 | + } |
| 78 | + |
| 79 | + if (!visit[q]) { |
| 80 | + res.push(trees[q]) |
| 81 | + visit[q] = true |
| 82 | + } |
| 83 | + |
| 84 | + p = q |
| 85 | + } while (p !== leftMost) |
| 86 | + |
| 87 | + return res |
| 88 | +} |
| 89 | +``` |
| 90 | + |
| 91 | +### Graham 算法 |
| 92 | + |
| 93 | +```ts |
| 94 | +/** |
| 95 | + * Graham 算法 |
| 96 | + * @desc 时间复杂度 O(NlogN) 空间复杂度 O(N) |
| 97 | + * @param trees |
| 98 | + * @returns |
| 99 | + */ |
| 100 | +export function outerTrees2(trees: number[][]): number[][] { |
| 101 | + const len = trees.length |
| 102 | + if (len < 4) |
| 103 | + return trees |
| 104 | + |
| 105 | + let bottom = 0 |
| 106 | + // 找到 y 最小的点 bottom |
| 107 | + for (let i = 0; i < len; i++) { |
| 108 | + if (trees[i][1] < trees[bottom][i]) |
| 109 | + bottom = i |
| 110 | + } |
| 111 | + |
| 112 | + const swap = (trees: number[][], i: number, j: number) => { |
| 113 | + [ |
| 114 | + trees[i][0], |
| 115 | + trees[i][1], |
| 116 | + trees[j][0], |
| 117 | + trees[j][1], |
| 118 | + ] |
| 119 | + = [ |
| 120 | + trees[j][0], |
| 121 | + trees[j][1], |
| 122 | + trees[i][0], |
| 123 | + trees[i][1], |
| 124 | + ] |
| 125 | + return trees |
| 126 | + } |
| 127 | + |
| 128 | + const cross = (p: number[], q: number[], r: number[]) => (q[1] - p[1]) * (r[0] - q[0]) - (q[0] - p[0]) * (r[1] - q[1]) |
| 129 | + |
| 130 | + const distance = (p: number[], q: number[]) => (p[0] - q[0]) * (p[0] - q[0]) + (p[1] - q[1]) * (p[1] - q[1]) |
| 131 | + |
| 132 | + trees = swap(trees, bottom, 0) |
| 133 | + |
| 134 | + // 以 bottom 原点，按照极坐标的角度大小进行排序 |
| 135 | + trees.sort((a, b) => { |
| 136 | + const diff = cross(trees[0], a, b) - cross(trees[0], b, a) |
| 137 | + return diff === 0 ? distance(trees[0], a) - distance(trees[0], b) : diff > 0 ? 1 : -1 |
| 138 | + }) |
| 139 | + |
| 140 | + // 对于凸包最后且在同一条直线的元素按照距离从小到大进行排序 |
| 141 | + let r = len - 1 |
| 142 | + while (r > 0 && cross(trees[0], trees[len - 1], trees[r]) === 0) |
| 143 | + r-- |
| 144 | + |
| 145 | + for (let l = r + 1, h = len - 1; l < h; l++, h--) |
| 146 | + trees = swap(trees, l, h) |
| 147 | + |
| 148 | + const stack = [trees[0], trees[1]] |
| 149 | + for (let i = 2; i < len; i++) { |
| 150 | + let top = stack.pop()! |
| 151 | + // 如果当前元素与栈顶的两个元素构成的向量顺时针旋转，则弹出栈顶元素 |
| 152 | + while (cross(stack[stack.length - 1], top, trees[i]) > 0) |
| 153 | + top = stack.pop()! |
| 154 | + |
| 155 | + stack.push(top) |
| 156 | + stack.push(trees[i]) |
| 157 | + } |
| 158 | + return stack |
| 159 | +} |
| 160 | +``` |
| 161 | + |
| 162 | +### Andrew 算法 |
| 163 | + |
| 164 | +```ts |
| 165 | +/** |
| 166 | + * Andrew 算法 |
| 167 | + * @desc 时间复杂度 O(NlogN) 空间复杂度 O(N) |
| 168 | + * @param trees |
| 169 | + * @returns |
| 170 | + */ |
| 171 | +export function outerTrees3(trees: number[][]): number[][] { |
| 172 | + const len = trees.length |
| 173 | + if (len < 4) |
| 174 | + return trees |
| 175 | + |
| 176 | + // 按照 x 大小进行排序，如果 x 相同，则按照 y 的大小进行排序 |
| 177 | + trees.sort((a, b) => { |
| 178 | + if (a[0] === b[0]) |
| 179 | + return a[1] - b[1] |
| 180 | + |
| 181 | + return a[0] - b[0] |
| 182 | + }) |
| 183 | + |
| 184 | + const hull = [] |
| 185 | + const used = new Array(len).fill(false) |
| 186 | + |
| 187 | + // hull[0] 需要入栈两次，不进行标记 |
| 188 | + hull.push(0) |
| 189 | + |
| 190 | + const cross = (p: number[], q: number[], r: number[]) => (q[0] - p[0]) * (r[1] - q[1]) - (q[1] - p[1]) * (r[0] - q[0]) |
| 191 | + |
| 192 | + // 求出凸包的下半部分 |
| 193 | + for (let i = 1; i < len; i++) { |
| 194 | + while (hull.length > 1 && cross(trees[hull[hull.length - 2]], trees[hull[hull.length - 1]], trees[i]) < 0) { |
| 195 | + used[hull[hull.length - 1]] = false |
| 196 | + hull.pop() |
| 197 | + } |
| 198 | + used[i] = true |
| 199 | + hull.push(i) |
| 200 | + } |
| 201 | + |
| 202 | + const m = hull.length |
| 203 | + // 求出凸包的上半部分 |
| 204 | + for (let i = len - 2; i >= 0; i--) { |
| 205 | + if (!used[i]) { |
| 206 | + while (hull.length > m && cross(trees[hull[hull.length - 2]], trees[hull[hull.length - 1]], trees[i]) < 0) { |
| 207 | + used[hull[hull.length - 1]] = false |
| 208 | + hull.pop() |
| 209 | + } |
| 210 | + used[i] = true |
| 211 | + hull.push(i) |
| 212 | + } |
| 213 | + } |
| 214 | + |
| 215 | + // hull[0] 同时参与凸包的上半部分检测，需要去掉重复的 hull[0] |
| 216 | + hull.pop() |
| 217 | + |
| 218 | + const size = hull.length |
| 219 | + const res: number[][] = [] |
| 220 | + for (let i = 0; i < size; i++) |
| 221 | + res[i] = trees[hull[i]] |
| 222 | + |
| 223 | + return res |
| 224 | +} |
| 225 | +``` |
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