0228. Summary Ranges (E)

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"]
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.

题意

给定一个升序数组，将其中每一个相邻元素连续的子数组以区间字符串的形式表示，要求返回这样的区间字符串的汇总。

思路

用两指针分别指向符合条件的子数组的两端，按照题目要求进行处理。

代码实现

Java

class Solution {
    public List<String> summaryRanges(int[] nums) {
        List<String> ans = new ArrayList<>();
        if (nums.length != 0) {
            int start = nums[0];
            int end = nums[0];
            for (int i = 1; i < nums.length; i++) {
                if (nums[i] == nums[i - 1] + 1) {
                    // 扩大子数组长度
                    end++;
                } else {
                    // 找到一个子数组，将其转化为区间字符串
                    String range = start == end ? start + "" : start + "->" + end;
                    ans.add(range);
                    start = nums[i];
                    end = nums[i];
                }
            }
            // 还需要再进行一次处理
            String range = start == end ? start + "" : start + "->" + end;
            ans.add(range);
        }
        return ans;
    }
}

JavaScript

/**
 * @param {number[]} nums
 * @return {string[]}
 */
var summaryRanges = function (nums) {
  if (!nums.length) {
    return []
  }
  let ans = []
  let start = nums[0]
  for (let i = 1; i < nums.length; i++) {
    if (nums[i] > nums[i - 1] + 1) {
      ans.push(nums[i - 1] == start ? start + '' : start + '->' + nums[i - 1])
      start = nums[i]
    }
  }
  ans.push(nums[nums.length - 1] == start ? start + '' : start + '->' + nums[nums.length - 1])
  return ans
}