| 62 | |
| 63 | // 数组手搓双向链表的实现 |
| 64 | public static void near2() { |
| 65 | for (int i = 1; i <= n; i++) { |
| 66 | rank[i][0] = i; |
| 67 | rank[i][1] = arr[i]; |
| 68 | } |
| 69 | Arrays.sort(rank, 1, n + 1, (a, b) -> a[1] - b[1]); |
| 70 | rank[0][0] = 0; |
| 71 | rank[n + 1][0] = 0; |
| 72 | for (int i = 1; i <= n; i++) { |
| 73 | last[rank[i][0]] = rank[i - 1][0]; |
| 74 | next[rank[i][0]] = rank[i + 1][0]; |
| 75 | } |
| 76 | for (int i = 1; i <= n; i++) { |
| 77 | to1[i] = 0; |
| 78 | dist1[i] = 0; |
| 79 | to2[i] = 0; |
| 80 | dist2[i] = 0; |
| 81 | update(i, last[i]); |
| 82 | update(i, last[last[i]]); |
| 83 | update(i, next[i]); |
| 84 | update(i, next[next[i]]); |
| 85 | delete(i); |
| 86 | } |
| 87 | } |
| 88 | |
| 89 | // i位置右侧的j位置 |
| 90 | // 看看能不能更新i右侧的最近或者次近 |