LeetCode_335 路径交叉（Java版）

只有如下三种情况：

    /*               i-2
    case 1 : i-1┌─┐
                └─┼─>i
                 i-3

                    i-2
    case 2 : i-1 ┌────┐
                 └─══>┘i-3
                 i  i-4      (i overlapped i-4)

    case 3 :    i-4
               ┌──┐
               │i<┼─┐
            i-3│ i-5│i-1
               └────┘
                i-2

*/

public class Solution_335 {
    public boolean isSelfCrossing(int[] distance) {
        for (int i = 3; i < distance.length; i++) {
            if (distance[i] >= distance[i-2] && distance[i-1] <= distance[i-3]){
                return true;
            }
            if (i == 4 && distance[1] == distance[3] && distance[0] + distance[4] == distance[2]){
                return true;
            }
            if (i >= 5 && distance[i-1] >= distance[i-3]-distance[i-5] && distance[i-1] <= distance[i-3] &&
                    distance[i] >= distance[i-2] - distance[i-4] && distance[i-2] >= distance[i-4]){
                return true;
            }

        }
        return false;
    }

}

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来源： https://www.cnblogs.com/Tianhaoblog/p/15483865.html