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[leetcode]summary ranges and missing ranges

admin 11月 11, 2020 0

题目描述

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”].

代码

public class Solution{
	
    //使用for循环遍历数组，对每一个元素不断与右边的元素相比较，如果相差为1则还是同一个连续区间，
    //下标i++，直到与右边元素之差不唯一则找到了一个连续区间
    public List<String> summaryRanges(int[] nums) {
        List<String> res = new ArrayList<>();
        if(nums == null || nums.length == 0) return res;
        for(int i = 0; i < nums.length; i++){
            int temp = nums[i];
            while(i+1 < nums.length && nums[i+1] - nums[i] == 1) i++;
            
            if(nums[i] != temp){
                res.add(temp + "->" + nums[i]);
            }else{
                res.add(temp+"");
            }
        }
        
        return res;
    }
}

题目描述

Given a sorted integer array where the range of elements are [0, 99] inclusive, return its missing ranges.
For example, given [0, 1, 3, 50, 75], return [“2”, “4->49”, “51->74”, “76->99”]

You should be able to extend the above cases not only for the range [0,99], but any arbitrary range [start, end].

代码

public class Solution{
	public List<String> getMissingRanges(int[] nums, int start, int end){
        List<String> ranges = new ArrayList<>();
        
        int prev = start - 1;
        for(int i = 0; i <= nums.length; i++){
            int curr = (i == nums.length) ? end + 1 : nums[i];
            if(curr - prev >= 2){
                ranges.add(getRange(prev + 1, curr - 1));
            }
            prev = curr;
        }
        return ranges;
    }
    
    private String (int from, int to){
        return from == to ? String.valueOf(from) : from + "->" + to;
    }
}