LeetCode.228 Summary Ranges
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题目:
Given a sorted integer array without duplicates, return the summary of its ranges.
Example 1:
Input: [0,1,2,4,5,7]Output: ["0->2","4->5","7"]
Example 2:
Input: [0,2,3,4,6,8,9]Output: ["0","2->4","6","8->9"]
Credits:
Special thanks to @jianchao.li.fighter for adding this problem and creating all test cases.
class Solution { public List<String> summaryRanges(int[] nums) { //给定排好序的数组,其中连续的元素放在一组,非连续的单独存放,返回结果 //思路:对于每个元素和前一个元素进行匹配 //注意:最后需要对最后单个元素或者元素对进行判断 List<String> list=new ArrayList<>(); if(nums.length==0||nums==null) return list; int count=1; int temp=nums[0]; int start=nums[0]; for(int i=1;i<nums.length;i++){ if(nums[i]==temp+1){ //为连续的 temp=nums[i]; count++; }else { //不连续 if(count==1){ //只有一个元素 list.add(String.valueOf(start)); }else if(count>1){ //有多个元素 String str=String.valueOf(start)+"->"+String.valueOf(temp); list.add(str); //更新start和temp } start=nums[i]; temp=nums[i]; count=1; } } //防止:对最后的元素进行处理 if(count>1){ //说明还有元素对 String str=String.valueOf(start)+"->"+String.valueOf(temp); list.add(str); }else if(count==1){ //最后只有单个元素 list.add(String.valueOf(start)); } return list; }}
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