[LeetCode] Self Crossing
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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.
Example 1:
Given x = [2, 1, 1, 2],
┌───┐
│ │
└───┼──>
Return true (self crossing)
Example 2:
Given x = [1, 2, 3, 4],
┌──────┐
│ │
│
│
└────────────>
Return false (not self crossing)
Example 3:
Given x = [1, 1, 1, 1],
┌───┐
│ │
└───┼>
Return true (self crossing)
解题思路
相交可分为如下三种情况:
- 第四条线与第一条线相交
- 第五条线与第一条线相交或重叠
- 第六条线与第一条线相交
点Xi为给定的最后一个点。
实现代码
// Runtime: 1 ms
public class Solution {
public boolean isSelfCrossing(int[] x) {
int len = x.length;
if(len <= 3) return false;
for(int i = 3; i < len; i++) {
if(x[i] >= x[i-2] && x[i-1] <= x[i-3]) {
return true;
}
if(i >= 4) {
if(x[i-1] == x[i-3] && x[i] + x[i-4] >= x[i-2]) {
return true;
}
}
if(i >= 5) {
if(x[i-2] - x[i-4] >= 0 && x[i] >= x[i-2] - x[i-4] && x[i-1] >= x[i-3] - x[i-5] && x[i-1] <= x[i-3]) {
return true;
}
}
}
return false;
}
}
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