Summary Ranges (M)
題目
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.
題意
給定一個升序陣列，將其中每一個相鄰元素連續的子陣列以區間字串的形式表示，要求回傳這樣的區間字串的匯總，
思路
用兩指標分別指向符合條件的子陣列的兩端，按照題目要求進行處理，
代碼實作
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) {
// 擴大子陣列長度
end++;
} else {
// 找到一個子陣列，將其轉化為區間字串
String range = start == end ? start + "" : start + "->" + end;
ans.add(range);
start = nums[i];
end = nums[i];
}
}
// 還需要再進行一次處理
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
}
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