题目:

Given a sorted integer array where the range of elements are [lowerupper] inclusive, return its missing ranges.

For example, given [0, 1, 3, 50, 75]lower = 0 and upper = 99, return ["2", "4->49", "51->74", "76->99"].

链接: http://leetcode.com/problems/missing-ranges/

题解:

查找丢失的range。这道题目主要注意一些边界条件。数组为空时要加入lower或者lower -> upper。不为空时可以先做一个特殊处理lower--,这样可以在变例数组的时候用条件getRangeAsString(lower + 1,nums[i] - 1)来把lower包括进来。遍历完毕数组之后要比较此时的lower以及upper,再尝试update list。

Time Complexity - O(n), Space Complexity - O(n)。

public class Solution {
public List<String> findMissingRanges(int[] nums, int lower, int upper) {
List<String> res = new ArrayList<>();
if(nums == null)
return res;
if(nums.length == 0) {
res.add(getRangeAsString(lower, upper));
return res;
} lower--; //to include lower;
for(int i = 0; i < nums.length; i++) {
if(nums[i] - lower >= 2) // lower always <= nums[i]
res.add(getRangeAsString(lower + 1, nums[i] - 1));
lower = nums[i];
} if(upper > lower) // boundary condition, include upper
res.add(getRangeAsString(lower + 1, upper)); return res;
} private String getRangeAsString(int lower, int upper) {
if(lower == upper)
return String.valueOf(lower);
StringBuilder sb = new StringBuilder();
sb.append(lower);
sb.append("->");
sb.append(upper);
return sb.toString();
}
}

二刷:

方法跟一刷一样,但更新了一下边界条件的写法。也是建立了一个辅助方法来返回单个字符或者一个range。 遍历数组的时候,每次当num - lower >= 1时,我们进行结果集的添加操作,然后更新lower = num + 1。遍历完毕以后要再check一下 lower 和 upper的关系。

Java:

Time Complexity - O(n), Space Complexity - O(n)。

public class Solution {
public List<String> findMissingRanges(int[] nums, int lower, int upper) {
List<String> res = new ArrayList<>();
if (nums == null || lower > upper) return res;
for (int num : nums) {
if (num - lower >= 1) res.add(getRangeString(lower, num - 1));
lower = num + 1;
}
if (lower <= upper) res.add(getRangeString(lower, upper));
return res;
} private String getRangeString(int lower, int upper) {
if (lower == upper) return String.valueOf(lower);
StringBuilder sb = new StringBuilder();
sb.append(lower).append("->").append(upper);
return sb.toString();
}
}

测试:

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