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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.
You are given an array x of
n
positive numbers. You start at point(0,0)
and movesx[0]
metres to the north, thenx[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:
Example 2:
Example 3:
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:
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...
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...
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:
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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