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”]
Explanation: 0,1,2 form a continuous range; 4,5 form a continuous range.
Example 2:
Input: [0,2,3,4,6,8,9]
Output: [“0″,”2->4″,”6″,”8->9”]
Explanation: 2,3,4 form a continuous range; 8,9 form a continuous range.
难度：medium
题目：给定排序且无重复元素的整数数组，返回其连续元素的范围。
思路：用以变量记录连续元素的开始。
Runtime: 5 ms, faster than 7.57% of Java online submissions for Summary Ranges.Memory Usage: 37.5 MB, less than 5.02% of Java online submissions for Summary Ranges.
class Solution {
public List<String> summaryRanges(int[] nums) {
List<String> result = new ArrayList<>();
if (null == nums || nums.length < 1) {
return result;
}
for (int i = 1, start = 0; i <= nums.length; i++) {
if (i == nums.length || nums[i] – nums[i – 1] != 1) {
result.add((i – 1 == start) ? String.format(“%s”, nums[start])
: String.format(“%s->%s”, nums[start], nums[i – 1]));
start = i;
}
}
return result;
}
}