一个算法题--Self Crossing
You are given an array x of n
positive numbers. You start at point (0,0)
and moves x[0]
metres to the north, then x[1]
metres to the west, x[2]
metres to the south,x[3]
metres to the east and so on. In other words, after each move your direction changes counter-clockwise.
Write a one-pass algorithm with O(1)
extra space to determine, if your path crosses itself, or not.
Example1:
Example2:
Example3:
Answer:
1 public boolean isSelfCrossing(int[] x) { 2 if(x==null||x.length<=3){ 3 return false; 4 } 5 for(int i=3;i<x.length;i++){ 6 if(x[i]>=x[i-2]&&x[i-1]<=x[i-3]){ // case 1 7 return true; 8 }else if(i>=4&&x[i-1]==x[i-3]&&x[i]+x[i-4]>=x[i-2]){ // case 2 9 return true; 10 }else if(i>=5&&x[i-2]>x[i-4]&&x[i]+x[i-4]>=x[i-2]&&x[i-1]<x[i-3]&&x[i-1]+x[i-5]>=x[i-3]){ //case 3 11 return true; 12 } 13 } 14 return false; 15 }
posted on 2016-03-02 15:09 Yanspecial 阅读(301) 评论(0) 编辑 收藏 举报