LeetCode 335. Self Crossing（贪吃蛇）
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原题网址：https://leetcode.com/problems/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.
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)方法：模拟移动的过程来判断，是否交叉跟最近6个点所连成的线段有关。
public class Solution { public boolean isSelfCrossing(int[] x) { if (x == null || x.length < 4) return false; int[] rows = new int[6]; int[] cols = new int[6]; for(int i=0; i<x.length; i++) { int nrow = rows[5], ncol = cols[5]; switch (i%4) { case 0: nrow -= x[i]; break; case 1: ncol -= x[i]; break; case 2: nrow += x[i]; break; case 3: ncol += x[i]; break; } if (i >= 5) { switch (i%4) { case 0: if (cols[0] <= ncol && ncol <= cols[1] && rows[5] >= rows[1] && rows[1] >= nrow) return true; else break; case 1: if (rows[1] <= nrow && nrow <= rows[0] && cols[5] >= cols[1] && cols[1] >= ncol) return true; else break; case 2: if (cols[1] <= ncol && ncol <= cols[0] && rows[5] <= rows[1] && rows[1] <= nrow) return true; else break; case 3: if (rows[0] <= nrow && nrow <= rows[1] && cols[5] <= cols[1] && cols[1] <= ncol) return true; else break; } } if (i >= 3) { switch (i%4) { case 0: if (cols[3] <= ncol && ncol <= cols[2] && rows[5] <= rows[3] && rows[3] <= nrow) return true; else break; case 1: if (rows[2] <= nrow && nrow <= rows[3] && cols[5] >= cols[3] && cols[3] >= ncol) return true; else break; case 2: if (cols[2] <= ncol && ncol <= cols[3] && rows[5] <= rows[2] && rows[2] <= nrow) return true; else break; case 3: if (rows[3] <= nrow && nrow <= rows[2] && cols[5] <= cols[2] && cols[2] <= ncol) return true; else break; } } if (i==4 && ncol == 0 && rows[5] > 0 && nrow <= 0) return true; rows[0] = rows[1]; rows[1] = rows[2]; rows[2] = rows[3]; rows[3] = rows[4]; rows[4] = rows[5]; rows[5] = nrow; cols[0] = cols[1]; cols[1] = cols[2]; cols[2] = cols[3]; cols[3] = cols[4]; cols[4] = cols[5]; cols[5] = ncol; } return false; }}另一种实现：
public class Solution { public boolean isSelfCrossing(int[] x) { int[] row = new int[7]; int[] col = new int[7]; for(int i=0; i<x.length; i++) { for(int j=0; j<6; j++) { row[j]=row[j+1]; col[j]=col[j+1]; } switch (i % 4) { case 0: row[6] = row[5] - x[i]; col[6] = col[5]; if (i >= 8 && col[0] <= col[5] && col[5] <= col[1] && row[6] <= row[0] && row[0] <= row[5]) return true; if (i >= 4 && col[3] <= col[5] && col[5] <= col[2] && row[6] <= row[2] && row[2] <= row[5]) return true; if (i == 4 && col[6] == 0 && row[6] <= 0) return true; break; case 1: row[6] = row[5]; col[6] = col[5] - x[i]; if (i >= 5 && col[6] <= col[0] && col[0] <= col[5] && row[1] <= row[5] && row[5] <= row[0]) return true; if (i >= 5 && col[6] <= col[2] && col[2] <= col[5] && row[2] <= row[5] && row[5] <= row[3]) return true; break; case 2: row[6] = row[5] + x[i]; col[6] = col[5]; if (i >= 6 && col[1] <= col[5] && col[5] <= col[0] && row[5] <= row[0] && row[0] <= row[6]) return true; if (i >= 6 && col[2] <= col[5] && col[5] <= col[3] && row[5] <= row[2] && row[2] <= row[6]) return true; break; case 3: row[6] = row[5]; col[6] = col[5] + x[i]; if (i >= 7 && col[5] <= col[0] && col[0] <= col[6] && row[0] <= row[5] && row[5] <= row[1]) return true; if (i >= 3 && col[5] <= col[2] && col[2] <= col[6] && row[3] <= row[5] && row[5] <= row[2]) return true; break; } } return false; }}另一种实现：
public class Solution { public boolean isSelfCrossing(int[] x) { int[] px = new int[6]; int[] py = new int[6]; int ny = 0; int nx = 0; for(int i = 0; i < x.length; i++) { switch (i % 4) { case 0: ny -= x[i]; break; case 1: nx -= x[i]; break; case 2: ny += x[i]; break; case 3: nx += x[i]; break; } if (i == 3) { if (px[3] <= px[0] && px[0] <= nx && py[1] <= ny && ny <= py[0]) return true; } else if (i == 4) { if (px[2] <= nx && nx <= px[1] && ny <= py[2] && py[2] <= py[4]) return true; if (nx == px[1] && ny <= py[0] && py[0] <= py[4]) return true; } else if (i > 4) { switch (i % 4) { case 0: if (px[0] <= nx && nx <= px[1] && ny <= py[1] && py[1] <= py[5]) return true; if (px[3] <= nx && nx <= px[2] && ny <= py[3] && py[3] <= py[5]) return true; break; case 1: if (nx <= px[1] && px[1] <= px[5] && py[1] <= ny && ny <= py[0]) return true; if (nx <= px[3] && px[3] <= px[5] && py[2] <= ny && ny <= py[3]) return true; break; case 2: if (px[1] <= nx && nx <= px[0] && py[5] <= py[1] && py[1] <= ny) return true; if (px[2] <= nx && nx <= px[3] && py[5] <= py[3] && py[3] <= ny) return true; break; case 3: if (px[5] <= px[1] && px[1] <= nx && py[0] <= ny && ny <= py[1]) return true; if (px[5] <= px[3] && px[3] <= nx && py[3] <= ny && ny <= py[2]) return true; break; } } if (i < 5) { px[i + 1] = nx; py[i + 1] = ny; } else { for(int j = 0; j < 5; j++) { px[j] = px[j + 1]; py[j] = py[j + 1]; } px[5] = nx; py[5] = ny; } } return false; }}有更巧妙的方法，参考：https://leetcode.com/discuss/88054/java-oms-with-explanation
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