Leetcode Missing Ranges

最新推荐文章于 2021-05-09 13:53:21 发布

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最新推荐文章于 2021-05-09 13:53:21 发布

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分类专栏： Leetcode Leetcode 重启

本文链接：https://blog.youkuaiyun.com/u011472124/article/details/51731754

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本文介绍了一个中等难度的算法问题，即如何找出指定范围内一个已排序整数数组所缺失的区间。通过一个示例说明了如何从给定的数组及范围中找出所有未出现的连续区间，并提供了一个Java实现方案。

Given a sorted integer array where the range of elements are [lower, upper] 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"].

Difficulty: Medium

public class Solution {
    public String genString(int start, int end){
        if(start == end)
            return String.valueOf(start);
        else{
            return String.valueOf(start) + "->" + String.valueOf(end);
        }
    }
    public List<String> findMissingRanges(int[] nums, int lower, int upper) {
        List<String> list = new ArrayList<String>();
        if(nums.length == 0){
            list.add(genString(lower, upper));
            return list;
        }
        int left = lower, right = -1;
        
        for(int i = 0; i < nums.length; i++){
            right = nums[i];
            if(i == 0){
                if(left == right){
                    ;
                }
                else if(left - right == -1){
                    list.add(String.valueOf(left));
                }
                else{
                    list.add(String.valueOf(left) + "->" + String.valueOf(right - 1));
                }
                left = right;
            }
            else{
                if(left == right || left - right == -1){
                    ;
                }
                else{
                    list.add(genString(left + 1, right - 1));
                }
                left = right;
            }
        }
        if(left == upper){
            ;
        }
        else if(left - upper == -1){
            list.add(String.valueOf(upper));
        }
        else{
            list.add(String.valueOf(left + 1) + "->" + String.valueOf(upper));
        }
        return list;
    }
}