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[LeetCode] 335. Self Crossing #335

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grandyang opened this issue May 30, 2019 · 0 comments
Open

[LeetCode] 335. Self Crossing #335

grandyang opened this issue May 30, 2019 · 0 comments

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@grandyang
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grandyang commented May 30, 2019

 

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)

这道题给了我们一个一位数组,每个数字是个移动量,按照上左下右的顺序来前进每一个位移量,问我们会不会和之前的轨迹相交,而且限定了常量的空间复杂度,我立马想到了贪吃蛇游戏,但是这条蛇不会自动向前移动哈。言归正传,这题我不会,参考的网上大神们的解法,实际上相交的情况只有以下三种情况:

     x(1)
    ┌───┐
x(2)│   │x(0)
    └───┼──>
    x(3)│

第一类是第四条边和第一条边相交的情况,需要满足的条件是第一条边大于等于第三条边,第四条边大于等于第二条边。同样适用于第五条边和第二条边相交,第六条边和第三条边相交等等,依次向后类推的情况...

 

      x(1)
    ┌──────┐
    │      │x(0)
x(2)│      ^
    │      │x(4)
    └──────│
      x(3)

第二类是第五条边和第一条边重合相交的情况,需要满足的条件是第二条边和第四条边相等,第五条边大于等于第三条边和第一条边的差值,同样适用于第六条边和第二条边重合相交的情况等等依次向后类推...

 

      x(1)
    ┌──────┐
    │      │x(0)
x(2)│     <│────│
    │       x(5)│x(4)
    └───────────│
        x(3)

第三类是第六条边和第一条边相交的情况,需要满足的条件是第四条边大于等于第二条边,第三条边大于等于第五条边,第五条边大于等于第三条边和第一条边的差值,第六条边大于等于第四条边和第二条边的差值,同样适用于第七条边和第二条边相交的情况等等依次向后类推...

那么根据上面的分析,我们不难写出代码如下:

class Solution {
public:
    bool isSelfCrossing(vector<int>& x) {
        for (int i = 3; i < x.size(); ++i) {
            if (x[i] >= x[i - 2] && x[i - 3] >= x[i - 1]) {
                return true;
            }
            if (i >= 4 && x[i-1] == x[i-3] && x[i] >= x[i-2] - x[i-4]) {
                return true;
            }
            if (i >= 5 && x[i-2] >= x[i-4] && x[i-3] >= x[i-1] && x[i-1] >= x[i-3] - x[i-5] && x[i] >= x[i-2] - x[i-4]) {
                return true;
            }
        }
        return false;
    }
};

 

参考资料:

https://leetcode.com/discuss/88054/java-oms-with-explanation

https://leetcode.com/discuss/88196/re-post-2-o-n-c-0ms-solutions

 

LeetCode All in One 题目讲解汇总(持续更新中...)

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