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Lintcode: Nuts & Bolts Problem
Given a set of n nuts of different sizes and n bolts of different sizes. There is a one-one mapping between nuts and bolts. Comparison of a nut to another nut or a bolt to another bolt is not allowed. It means nut can only be compared with bolt and bolt can only be compared with nut to see which one is bigger/smaller. We will give you a compare function to compare nut with bolt.
其实是一种特别顺序的排序, 所以就用quicksort-partition, 不过比较的元素和设计互相patition, partition 的比较因为是有equals的, 所以得用三个指针遍历
/** * public class NBCompare { * public int cmp(String a, String b); * } * You can use compare.cmp(a, b) to compare nuts "a" and bolts "b", * if "a" is bigger than "b", it will return 1, else if they are equal, * it will return 0, else if "a" is smaller than "b", it will return -1. * When "a" is not a nut or "b" is not a bolt, it will return 2, which is not valid. */ public class Solution { /** * @param nuts: an array of integers * @param bolts: an array of integers * @param compare: a instance of Comparator * @return: nothing */ public void sortNutsAndBolts(String[] nuts, String[] bolts, NBComparator compare) { if (nuts == null || bolts == null) return; if (nuts.length != bolts.length) return; qsort(nuts, bolts, compare, 0, nuts.length - 1); } private void qsort(String[] nuts, String[] bolts, NBComparator compare, int l, int u) { if (l >= u) return; // find the partition index for nuts with bolts[u] int part_inx = partition(nuts, bolts[u], compare, l, u); // partition bolts with nuts[part_inx] partition(bolts, nuts[part_inx], compare, l, u); // qsort recursively qsort(nuts, bolts, compare, l, part_inx - 1); qsort(nuts, bolts, compare, part_inx + 1, u); } private int partition(String[] str, String pivot, NBComparator compare, int l, int u) { int low = l; int high = u; int i = low; while (i <= high) { if (compare.cmp(str[i], pivot) == -1 || compare.cmp(pivot, str[i]) == 1) { swap(str, i, low); i++; low++; } else if (compare.cmp(str[i], pivot) == 1 || compare.cmp(pivot, str[i]) == -1) { swap(str, i, high); high--; } else i++; } return low; } private void swap(String[] str, int l, int r) { String temp = str[l]; str[l] = str[r]; str[r] = temp; } }
考点: partition 的设计, pivot 的选择 , 另一方的排序通过nuts排好后的数组再排, partition 不仅 排序还返回了bolts的pivot
Lintcode: Nuts & Bolts Problem
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