228. Summary Ranges

题目：

Given a sorted integer array without duplicates, return the summary of its ranges.

For example, given [0,1,2,4,5,7], return ["0->2","4->5","7"].

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

题解：

总结Range。也是从头到尾走一遍。当nums[i] - 1> nums[i - 1]，我们处理之前的数字们。当遍历到最后一个元素的时候也要考虑如何处理。

Time Complexity - O(n)， Space Complexity - O(1)。

public class Solution {public List<String> summaryRanges(int[] nums) {List<String> res = new ArrayList<>();if(nums == null || nums.length == 0)return res;int lo = 0;StringBuilder sb = new StringBuilder();for(int i = 0; i < nums.length; i++) {if(i > 0 && (nums[i] - 1 > nums[i - 1])) {if(lo == i - 1)res.add(Integer.toString(nums[lo]));else {res.add(sb.append(nums[lo]).append("->").append(nums[i - 1]).toString());sb.setLength(0);}lo = i;}if(i == nums.length - 1) {if(lo == i)res.add(Integer.toString(nums[lo]));elseres.add(sb.append(nums[lo]).append("->").append(nums[i]).toString());}}return res;}
}

二刷:

方法和一刷一样。我们主要使用一个变量lo来保存每个interval的左边界。
每次当 i > 0并且 nums[i] - 1 > nums[i]的时候，我们进行判断
1. 假如 lo = i - 1，那么我们只有一个数字，直接把这个数字加入到结果
2. 否则我们要把 nums[lo] + "->" + nums[i - 1]这个字符串加入到结果
更新 lo = i
当i = len - 1的时候，我们也要根据上面的逻辑判断一遍
最后返回结果.

Java:

可以使用StringBuilder来减少空间复杂度。

Time Complexity - O(n)， Space Complexity - O(1)。

public class Solution {public List<String> summaryRanges(int[] nums) {List<String> res = new ArrayList<>();if (nums == null || nums.length == 0) {return res;}int lo = 0, len = nums.length;for (int i = 0; i < len; i++) {if (i > 0 && (nums[i] - 1 > nums[i - 1])) {if (lo == i - 1) {res.add(nums[lo]  + "");} else {res.add(nums[lo]  + "->" + nums[i - 1]);}lo = i;}if (i == len - 1) {if (lo == len - 1) {res.add(nums[lo] + "");} else {res.add(nums[lo] + "->" + nums[len - 1]);}}}return res;}
}

三刷:

不知道上面在写什么...这回主要使用一个变量count和一个StringBuilder sb。遍历整个数组，当count = 0的时候，我们在sb中加入当前nums[i]。当nums[i] - nums[i - 1]时，我们增加count。否则，这时nums[i] - nums[i - 1] > 1，假如count > 1，则形成了一个range，我们在sb中append一个符号"->"，再append上一个数字，把sb输出到结果。否则count = 1，我们直接输出sb到结果。最后运行完毕时假如count仍然>0，我们做相应步骤，把sb输出到结果。

Java:

public class Solution {public List<String> summaryRanges(int[] nums) {List<String> res = new ArrayList<>();if (nums == null || nums.length == 0) return res;StringBuilder sb = new StringBuilder();int count = 0;for (int i = 0; i < nums.length; i++) {if (count == 0) {sb.append(nums[i]);count++;} else if (nums[i] - nums[i - 1] == 1) {count++;} else {if (count > 1) sb.append("->").append(nums[i - 1]);res.add(sb.toString());sb.setLength(0);sb.append(nums[i]);count = 1;}}if (count > 1) sb.append("->").append(nums[nums.length - 1]);res.add(sb.toString());return res;}
}

简化一下。分析的时候要考虑起始状态，过程分支以及结束时的边界条件。

public class Solution {public List<String> summaryRanges(int[] nums) {List<String> res = new ArrayList<>();if (nums == null || nums.length == 0) return res;StringBuilder sb = new StringBuilder(nums[0] + "");int count = 1;for (int i = 1; i < nums.length; i++) {if (nums[i] - nums[i - 1] == 1) {count++;} else {if (count > 1) sb.append("->").append(nums[i - 1]);res.add(sb.toString());sb.setLength(0);sb.append(nums[i]);count = 1;}}if (count > 1) sb.append("->").append(nums[nums.length - 1]);res.add(sb.toString());return res;}
}

Update:

Different logic. This time we use a sliding window. If we found out nums[i] > nums[i - 1] + 1, we need to add result to res. here if i - 1 != lo, we need to add a range into res, otherwise we need to add a single number into res. We also need to do double check while we finished running the loop.

public class Solution {public List<String> summaryRanges(int[] nums) {List<String> res = new ArrayList<>();if (nums == null || nums.length == 0) return res;StringBuilder sb = new StringBuilder();int lo = 0, len = nums.length;for (int i = 1; i < len; i++) {if (nums[i] > nums[i - 1] + 1) {if (i - 1 != lo) sb.append(nums[lo]).append("->").append(nums[i - 1]);else sb.append(nums[lo]);res.add(sb.toString());lo = i;sb.setLength(0);}}if (lo == len - 1) sb.append(nums[lo]);else sb.append(nums[lo]).append("->").append(nums[len - 1]);res.add(sb.toString());return res;}
}

Reference:

https://leetcode.com/discuss/42229/10-line-c-easy-understand

https://leetcode.com/discuss/42199/6-lines-in-python

https://leetcode.com/discuss/42342/idea-1-liner-group-by-number-index

https://leetcode.com/discuss/42290/accepted-java-solution-easy-to-understand

转载于:https://www.cnblogs.com/yrbbest/p/4996505.html

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