M

1528247314

tags: Array

给一串sorted list, 中间有缺数字, return 所有数字的range string (example 看题目)

#### Basic implementation

- 用一个list as the buffer to store candidates

- when: 1. end of nums; 2. not continuous integer => convert list to result

```

/*

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"].

Tags: Array

Similar Problems: (M) Missing Ranges

*/

/*

Thoughts: basic implementation, use a arraylist to catch candidates.

Detect condition, and return results.

*/

public class Solution {

public List<String> summaryRanges(int[] nums) {

List<String> rst = new ArrayList<>();

if (nums == null || nums.length == 0) {

return rst;

}

List<Integer> list = new ArrayList<>();

for (int i = 0; i < nums.length; i++) {

list.add(nums[i]);

if (i + 1 == nums.length || nums[i] + 1 != nums[i + 1]) {

if (list.size() == 1) {

rst.add(list.get(0) + "");

} else {

rst.add(list.get(0) + "->" + list.get(list.size() - 1));

}

list = new ArrayList<Integer>();

}

}

return rst;

}

}

//O(n)

```