LeetCode 228. Summary Ranges（归纳区间）

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原题网址：https://leetcode.com/problems/summary-ranges/

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"].

思路：对每个元素，如果与上一个区间的右端点相差1，则合并区间，否则新开一个区间。

public class Solution {    public List<String> summaryRanges(int[] nums) {        Range[] ranges = new Range[nums.length];        int size = 0;        for(int i=0; i<nums.length; i++) {            if (i > 0 && ranges[size-1].to + 1 == nums[i]) ranges[size-1].to ++;            else ranges[size++] = new Range(nums[i], nums[i]);        }        List<String> results = new ArrayList<String>();        for(int i=0; i<size; i++) {            if (ranges[i].from == ranges[i].to) results.add(Integer.toString(ranges[i].from));            else results.add(Integer.toString(ranges[i].from) + "->" + Integer.toString(ranges[i].to));        }        return results;    }}class Range {    int from, to;    Range(int from, int to) {        this.from = from;        this.to = to;    }}

另一种类似的方法：

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

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