435. 无重叠区间

题目链接：. - 力扣（LeetCode）

文档讲解：代码随想录

视频讲解：贪心算法，依然是判断重叠区间 | LeetCode：435.无重叠区间_哔哩哔哩_bilibili

状态：通过

代码实现： 

class Solution {
public:
    static bool cmp(vector<int>& a, vector<int>& b){
        return a[0] < b[0];
    }
    int eraseOverlapIntervals(vector<vector<int>>& intervals) {
        int cnt = 0;
        sort(intervals.begin(), intervals.end(), cmp);
        for(int i = 1; i < intervals.size(); i++){
            if(intervals[i][0] < intervals[i - 1][1]){
                cnt++;
                intervals[i][1] = min(intervals[i - 1][1], intervals[i][1]);
            }
        }
        return cnt;
    }
};

763.划分字母区间

题目链接：. - 力扣（LeetCode）

文档讲解：代码随想录

视频讲解：贪心算法，寻找最远的出现位置！ LeetCode：763.划分字母区间_哔哩哔哩_bilibili

状态：未通过

代码实现： 

class Solution {
public:
    vector<int> partitionLabels(string s) {
        int hash[27] = {0};
        for(int i = 0; i < s.size(); i++){
            hash[s[i] - 'a'] = i;
        }
        vector<int> result;
        int left = 0;
        int right = 0;
        for(int i = 0; i < s.size(); i++){
            right = max(right, hash[s[i] - 'a']);
            if(i == right){
                result.push_back(right - left + 1);
                left = right + 1;
            }
        }
        return result;
    }
};

56. 合并区间

题目链接：. - 力扣（LeetCode）

文档讲解：代码随想录

视频讲解：贪心算法，合并区间有细节！LeetCode：56.合并区间_哔哩哔哩_bilibili

状态：通过

代码实现： 

class Solution {
public:
    static bool cmp(vector<int>& a, vector<int>& b){
        return a[0] < b[0];
    }
    vector<vector<int>> merge(vector<vector<int>>& intervals) {
        sort(intervals.begin(), intervals.end(), cmp);
        int left, right;
        vector<vector<int>> result;
        for(int i = 1; i < intervals.size(); i++){
            if(intervals[i][0] <= intervals[i - 1][1]){
                left = intervals[i - 1][0];
                right = max(intervals[i - 1][1], intervals[i][1]);
                intervals[i][0] = left;
                intervals[i][1] = right;
            }
            else{
                result.push_back(intervals[i - 1]);
            }
        }
        result.push_back(intervals[intervals.size() - 1]);
        return result;
    }
};