标签：int xpl public 含义 ever result exist 题目 class
Given a circular array (the next element of the last element is the first element of the array), print the Next Greater Number for every element. The Next Greater Number of a number x is the first greater number to its traversing-order next in the array, which means you could search circularly to find its next greater number. If it doesn‘t exist, output -1 for this number.
Example 1:
Input: [1,2,1] Output: [2,-1,2] Explanation: The first 1‘s next greater number is 2;
The number 2 can‘t find next greater number;
The second 1‘s next greater number needs to search circularly, which is also 2.
Note: The length of given array won‘t exceed 10000.
题目含义：给一个循环数组，返回一个等长的数组，数组中的每一个元素是：它后面的第一个大于它的元素（如果后面没有就循环一遍到最前面找，直到循环了一圈为止），如果不存在这样的数，就返回-1～
1 class Solution { 2 private int getNextGreater(int[] nums,int begin,int end,int target) 3 { 4 if (begin>nums.length -1 || end>nums.length-1 || begin > end) return Integer.MIN_VALUE; 5 for (int i=begin;i<=end;i++) 6 { 7 if (nums[i] > target) return nums[i]; 8 } 9 return Integer.MIN_VALUE; 10 } 11 12 public int[] nextGreaterElements(int[] nums) { 13 int[] result = new int[nums.length]; 14 if (nums.length==0) return new int[]{}; 15 if (nums.length ==1) return new int[]{-1}; 16 for (int i=0;i<nums.length;i++) 17 { 18 int greaterLeft = Integer.MIN_VALUE; 19 int greaterRight = Integer.MIN_VALUE; 20 if (i==0) 21 { 22 greaterRight = getNextGreater(nums,1,nums.length-1,nums[i]); 23 }else if (i == nums.length-1) 24 { 25 greaterLeft = getNextGreater(nums,0,i-1,nums[i]); 26 }else 27 { 28 greaterRight = getNextGreater(nums,i+1,nums.length-1,nums[i]); 29 if (greaterRight==Integer.MIN_VALUE) 30 { 31 greaterLeft = getNextGreater(nums,0,i-1,nums[i]); 32 } 33 } 34 if (greaterRight==Integer.MIN_VALUE && greaterLeft==Integer.MIN_VALUE) result[i] =-1; 35 else result[i] = Math.max(greaterLeft,greaterRight); 36 } 37 return result; 38 } 39 }
标签：int xpl public 含义 ever result exist 题目 class
原文地址：http://www.cnblogs.com/wzj4858/p/7725860.html
