标签：content find ted i++ sort list for input put
Given m arrays, and each array is sorted in ascending order. Now you can pick up two integers from two different arrays (each array picks one) and calculate the distance. We define the distance between two integers a and b to be their absolute difference |a-b|. Your task is to find the maximum distance.
Example 1:
Input: [[1,2,3], [4,5], [1,2,3]] Output: 4 Explanation: One way to reach the maximum distance 4 is to pick 1 in the first or third array and pick 5 in the second array.
Note:
m arrays will be in the range of [2, 10000].m arrays will be in the range of [-10000, 10000].
1 public class Solution { 2 public int maxDistance(List<List<Integer>> arrays) { 3 int n = arrays.size(); 4 int[] small = new int[n]; 5 int[] large = new int[n]; 6 for (int i = 0; i < n; i++) { 7 List<Integer> array = arrays.get(i); 8 small[i] = array.get(0); 9 large[i] = array.get(array.size() - 1); 10 } 11 12 HashMap<Integer, Integer> smallMap = new HashMap<>(); 13 HashMap<Integer, Integer> largeMap = new HashMap<>(); 14 for (int i = 0; i < n; i++) { 15 smallMap.put(small[i], i); 16 largeMap.put(large[i], i); 17 } 18 19 Arrays.sort(small); 20 Arrays.sort(large); 21 22 int low = 0, high = n - 1; 23 if (smallMap.get(small[low]) == largeMap.get(large[high])) { 24 25 // choose the second smallest 26 int result1 = calculateABS(large[high], small[low]); 27 if (low < n - 1) { 28 result1 = calculateABS(large[high], small[low + 1]); 29 } 30 31 // choose the second largest 32 int result2 = calculateABS(large[high], small[low]); 33 if (high > 0) { 34 result2 = calculateABS(large[high - 1], small[low]); 35 } 36 37 return Math.max(result1, result2); 38 } 39 return calculateABS(large[high], small[low]); 40 } 41 42 private int calculateABS(int num1, int num2) { 43 return Math.abs(num1 - num2); 44 } 45 }
标签：content find ted i++ sort list for input put
原文地址：http://www.cnblogs.com/amazingzoe/p/7053683.html
