【算法日志】贪心算法刷题：重叠区问题(day31)

代码随想录刷题60Day

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目录

前言

无重叠区间（筛选区间）

划分字母区间（切割区间）

 合并区间

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前言

今日的重点是掌握重叠区问题。

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无重叠区间（筛选区间）

   int eraseOverlapIntervals(vector>& intervals) 
    {
        int count = 0;
        const int size = intervals.size();
        sort(intervals.begin(), intervals.end());
        for (int i = 0; i < size; ++i)
        {
            int right = intervals[i][1];
            int j;
            for (j = i + 1; j < size; ++j)
            {
                if (right > intervals[j][1])
                    right = intervals[j][1];
                else if (right <= intervals[j][0])
                    break;
            }
            i = j - 1;
            ++count;
        }
        return size - count;
    }

划分字母区间（切割区间）

   vector partitionLabels(string s) 
    {
        vector result;
        const int size = s.length();
        int front = 0, back = size - 1;
        while(front < size)
        {
            while (s[front] != s[back])
                --back;
            if (front < back)
                for (int i = front + 1; i <= back; ++i)
                {
                    while (s[i] == s[i - 1])++i;
                    if (i > back) break;
                    int j  = size - 1;
                    while (j > back && s[i] != s[j])
                        j--;
                    back = j;
                }
            result.push_back(back - front + 1);
            front = back + 1;
            back = size - 1;
        }
        return result;
    }

 合并区间

   vector> merge(vector>& intervals) 
    {
        vector> result;
        sort(intervals.begin(), intervals.end());
        int size = intervals.size();
        for (int i = 0; i < size; ++i)
        {
            int j;
            int right = intervals[i][1];
            for (j = i + 1; j < size; ++j)
            {
                if (right < intervals[j][0])
                    break;
                else
                    right = max(right, intervals[j][1]);
            }
            result.push_back({intervals[i][0], right});
            i = j - 1;
        }
        return result;
    }