228.Summary Ranges
Difficulty: Medium
Given a sorted integer array without duplicates, return the summary of its ranges.
For example, given [0,1,2,4,5,7], return ["0->2","4->5","7"].
1. 解法，最简单，遍历判断。
public class Solution {public List<String> summaryRanges(int[] nums) {int length = nums.length;List<String> result = new ArrayList<String>();if (length == 0) {return result;}int start = nums[0];int theoreticalValue = start + 1;for (int index = 0; index < length; index ++) {if (index == length - 1) {if (nums[index] == (theoreticalValue - 1) && start != nums[index]) {String item = start + "->" + nums[index];result.add(item);} else {String item = "" + nums[index];result.add(item);}} else if(nums[index + 1] != theoreticalValue) {if (start == nums[index]) {String item = "" + nums[index];result.add(item);start = nums[index + 1];theoreticalValue = start;} else {String item = start + "->" + nums[index];result.add(item);start = nums[index + 1];theoreticalValue = start;}}theoreticalValue ++;}return result;}
}
虽然这个算法简单，当是并没有很好的效率，至打败了5.3%的java提交，这个题目本身不难，但是基本分成三类 ，好像是在第二类？好吧，其实用Python有更加就简单的方法。