[Leetcode]Summary Ranges

最新推荐文章于 2024-10-06 20:57:13 发布

dyllanzhou

最新推荐文章于 2024-10-06 20:57:13 发布

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分类专栏： algorithm 文章标签： leetcode 算法

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此博客展示了如何处理一个不重复的整数数组，并将其转换为一系列区间表示。通过迭代数组并跟踪连续元素，作者提供了两种方法来构建区间：一种是使用循环，另一种是递归方法。最终结果以字符串形式呈现，每个区间由起始和结束点表示，或者单独表示单一元素。

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"].

class Solution {
public:
    string itos(int num){
        stringstream ss;
        ss << num;
        return ss.str();
    }
    //[start,end]
    string rangeBuilder(int start,int end){
        if(start < end){
            return itos(start) + "->" + itos(end);
        }else
            return itos(start);
    }
    /*algorithm
        use start,end to track the window of range
        if nums[i] != nums[i-1]+1,reset start pointer
    */
    vector<string> summaryRanges(vector<int>& nums) {
            int start,end;
            vector<string>result;
            for(int i = 0;i < nums.size();i++){
                if(i == 0)start = end = i;
                else if(nums[i] != (nums[i-1] + 1)){
                    result.push_back(rangeBuilder(start,end));
                    start = end = i;
                }else end = i;
            }
            if((end + 1) == nums.size())
                result.push_back(rangeBuilder(start,end));
            return result;
    }
};

    /*recursive version 
    */ 
    //[start,end)  [0,1,2,4,5,7] 
    void summaryRangesSub(vector<int>&nums,int start,int end){ 
        if(start >= end)return; 
        int i = start; 
        while((i+1) < end && (nums[i]+1) == nums[i+1])++i; 
        result.push_back(rangeBuilder(nums,start,i)); 
        summaryRangesSub(nums,i + 1,end); 
    } 
    vector<string>result; 
    vector<string> summaryRanges(vector<int>& nums) { 
            summaryRangesSub(nums,0,nums.size()); 
            return result;