目录
1，题目描述
英文描述
中文描述
2，解题思路
3，AC代码
C++
Java
4，解题过程
第一博
1，题目描述
原题链接228. 汇总区间
英文描述
You are given a sorted unique integer array nums.
Return the smallest sorted list of ranges that cover all the numbers in the array exactly. That is, each element of nums is covered by exactly one of the ranges, and there is no integer x such that x is in one of the ranges but not in nums.
Each range [a,b] in the list should be output as:
"a->b" if a != b
"a" if a == b
Example 1:
Input: nums = [0,1,2,4,5,7]
Output: ["0->2","4->5","7"]
Explanation: The ranges are:
[0,2] --> "0->2"
[4,5] --> "4->5"
[7,7] --> "7"
Example 2:
Input: nums = [0,2,3,4,6,8,9]
Output: ["0","2->4","6","8->9"]
Explanation: The ranges are:
[0,0] --> "0"
[2,4] --> "2->4"
[6,6] --> "6"
[8,9] --> "8->9"
Example 3:
Input: nums = []
Output: []
Example 4:
Input: nums = [-1]
Output: ["-1"]
Example 5:
Input: nums = [0]
Output: ["0"]
Constraints:
0 <= nums.length <= 20
-231 <= nums[i] <= 231 - 1
All the values of nums are unique.
中文描述
给定一个无重复元素的有序整数数组 nums 。
返回 恰好覆盖数组中所有数字 的 最小有序 区间范围列表。也就是说，nums 的每个元素都恰好被某个区间范围所覆盖，并且不存在属于某个范围但不属于 nums 的数字 x 。
列表中的每个区间范围 [a,b] 应该按如下格式输出：
"a->b" ，如果 a != b
"a" ，如果 a == b
示例 1：
输入：nums = [0,1,2,4,5,7]
输出：["0->2","4->5","7"]
解释：区间范围是：
[0,2] --> "0->2"
[4,5] --> "4->5"
[7,7] --> "7"
示例 2：
输入：nums = [0,2,3,4,6,8,9]
输出：["0","2->4","6","8->9"]
解释：区间范围是：
[0,0] --> "0"
[2,4] --> "2->4"
[6,6] --> "6"
[8,9] --> "8->9"
示例 3：
输入：nums = []
输出：[]
示例 4：
输入：nums = [-1]
输出：["-1"]
示例 5：
输入：nums = [0]
输出：["0"]
提示：
0 <= nums.length <= 20
-231 <= nums[i] <= 231 - 1
nums 中的所有值都 互不相同
来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/summary-ranges
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。
2，解题思路
直接遍历，找到一段连续的整数，将区间两端转换为字符串进行拼接。（这居然是中等难度。。。）
3，AC代码
C++
class Solution {
public:
vector<string> summaryRanges(vector<int>& nums) {
vector<string> ans;
for(int index = 0; index < nums.size(); index++) {
int start = nums[index];
while(index + 1 < nums.size() && nums[index] + 1 == nums[index + 1]) {
index++;
}
if(start != nums[index])
ans.push_back(to_string(start) + "->" + to_string(nums[index]));
else
ans.push_back(to_string(start));
}
return ans;
}
};
Java
class Solution {
public List<String> summaryRanges(int[] nums) {
List<String> ans = new ArrayList<>();
for(int index = 0; index < nums.length; index++) {
int start = nums[index];
while(index + 1 < nums.length && nums[index] + 1 == nums[index + 1]) {
index++;
}
if(start != nums[index])
ans.add(start + "->" + nums[index]);
else
ans.add(start + "");
}
return ans;
}
}
贴一个100%的Java代码，暂时没看懂，以后再研究吧
class Solution {
public List<String> summaryRanges(int[] nums) {
if (nums == null || nums.length == 0) {
return new ArrayList();
}
List<String> strs = new ArrayList(nums.length);
int startPosition = 0;
int i = 1;
StringBuilder builder=new StringBuilder();
for (; i < nums.length; i++) {
if (nums[i] != nums[i - 1] + 1) {
if (startPosition == i - 1) {
strs.add(String.valueOf(nums[startPosition]));
} else {
builder.append(nums[startPosition]);
builder.append("->");
builder.append(nums[i - 1]);
strs.add(builder.toString());
builder.delete(0,builder.length());
}
startPosition = i;
}
}
if (startPosition == i - 1) {
strs.add(String.valueOf(nums[startPosition]));
} else {
builder.append(nums[startPosition]);
builder.append("->");
builder.append(nums[i - 1]);
strs.add(builder.toString());
builder.delete(0,builder.length());
}
return strs;
}
}
4，解题过程
第一博
for循环遍历，拼接相邻的区间
算法很简单，但是为啥结果这样呢？(；′⌒`)
