|
47 | 47 |
|
48 | 48 | <!-- 这里可写通用的实现逻辑 -->
|
49 | 49 |
|
| 50 | +**方法一:数学(向量叉积)** |
| 51 | + |
| 52 | +假设当前连续的三个顶点分别为 $p_1, p_2, p_3$,我们可以计算向量 $\overrightarrow{p_1p_2}$ 和 $\overrightarrow{p_1p_3}$ 的叉积,记为 $cur$。如果 $cur$ 的方向与之前的 $pre$ 方向不一致,说明多边形不是凸多边形。否则,我们更新 $pre = cur$,继续遍历下一个顶点。 |
| 53 | + |
| 54 | +遍历结束,如果没有发现不一致的情况,说明多边形是凸多边形。 |
| 55 | + |
| 56 | +时间复杂度 $O(n)$,其中 $n$ 是顶点的数量。空间复杂度 $O(1)$。 |
| 57 | + |
50 | 58 | <!-- tabs:start -->
|
51 | 59 |
|
52 | 60 | ### **Python3**
|
53 | 61 |
|
54 | 62 | <!-- 这里可写当前语言的特殊实现逻辑 -->
|
55 | 63 |
|
56 | 64 | ```python
|
57 |
| - |
| 65 | +class Solution: |
| 66 | + def isConvex(self, points: List[List[int]]) -> bool: |
| 67 | + n = len(points) |
| 68 | + pre = cur = 0 |
| 69 | + for i in range(n): |
| 70 | + x1 = points[(i + 1) % n][0] - points[i][0] |
| 71 | + y1 = points[(i + 1) % n][1] - points[i][1] |
| 72 | + x2 = points[(i + 2) % n][0] - points[i][0] |
| 73 | + y2 = points[(i + 2) % n][1] - points[i][1] |
| 74 | + cur = x1 * y2 - x2 * y1 |
| 75 | + if cur != 0: |
| 76 | + if cur * pre < 0: |
| 77 | + return False |
| 78 | + pre = cur |
| 79 | + return True |
58 | 80 | ```
|
59 | 81 |
|
60 | 82 | ### **Java**
|
61 | 83 |
|
62 | 84 | <!-- 这里可写当前语言的特殊实现逻辑 -->
|
63 | 85 |
|
64 | 86 | ```java
|
| 87 | +class Solution { |
| 88 | + public boolean isConvex(List<List<Integer>> points) { |
| 89 | + int n = points.size(); |
| 90 | + long pre = 0, cur = 0; |
| 91 | + for (int i = 0; i < n; ++i) { |
| 92 | + var p1 = points.get(i); |
| 93 | + var p2 = points.get((i + 1) % n); |
| 94 | + var p3 = points.get((i + 2) % n); |
| 95 | + int x1 = p2.get(0) - p1.get(0); |
| 96 | + int y1 = p2.get(1) - p1.get(1); |
| 97 | + int x2 = p3.get(0) - p1.get(0); |
| 98 | + int y2 = p3.get(1) - p1.get(1); |
| 99 | + cur = x1 * y2 - x2 * y1; |
| 100 | + if (cur != 0) { |
| 101 | + if (cur * pre < 0) { |
| 102 | + return false; |
| 103 | + } |
| 104 | + pre = cur; |
| 105 | + } |
| 106 | + } |
| 107 | + return true; |
| 108 | + } |
| 109 | +} |
| 110 | +``` |
| 111 | + |
| 112 | +### **C++** |
| 113 | + |
| 114 | +```cpp |
| 115 | +class Solution { |
| 116 | +public: |
| 117 | + bool isConvex(vector<vector<int>>& points) { |
| 118 | + int n = points.size(); |
| 119 | + long long pre = 0, cur = 0; |
| 120 | + for (int i = 0; i < n; ++i) { |
| 121 | + int x1 = points[(i + 1) % n][0] - points[i][0]; |
| 122 | + int y1 = points[(i + 1) % n][1] - points[i][1]; |
| 123 | + int x2 = points[(i + 2) % n][0] - points[i][0]; |
| 124 | + int y2 = points[(i + 2) % n][1] - points[i][1]; |
| 125 | + cur = 1L * x1 * y2 - x2 * y1; |
| 126 | + if (cur != 0) { |
| 127 | + if (cur * pre < 0) { |
| 128 | + return false; |
| 129 | + } |
| 130 | + pre = cur; |
| 131 | + } |
| 132 | + } |
| 133 | + return true; |
| 134 | + } |
| 135 | +}; |
| 136 | +``` |
65 | 137 |
|
| 138 | +### **Go** |
| 139 | +
|
| 140 | +```go |
| 141 | +func isConvex(points [][]int) bool { |
| 142 | + n := len(points) |
| 143 | + pre, cur := 0, 0 |
| 144 | + for i := range points { |
| 145 | + x1 := points[(i+1)%n][0] - points[i][0] |
| 146 | + y1 := points[(i+1)%n][1] - points[i][1] |
| 147 | + x2 := points[(i+2)%n][0] - points[i][0] |
| 148 | + y2 := points[(i+2)%n][1] - points[i][1] |
| 149 | + cur = x1*y2 - x2*y1 |
| 150 | + if cur != 0 { |
| 151 | + if cur*pre < 0 { |
| 152 | + return false |
| 153 | + } |
| 154 | + pre = cur |
| 155 | + } |
| 156 | + } |
| 157 | + return true |
| 158 | +} |
66 | 159 | ```
|
67 | 160 |
|
68 | 161 | ### **...**
|
|
0 commit comments