算法训练营38

题目1：435. 无重叠区间 - 力扣（LeetCode）

这道题其实跟上面那个像，我写的代码是遇到重复的就删除，不过没有AC过

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

题目2：763. 划分字母区间 - 力扣（LeetCode）

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

题目3：56. 合并区间 - 力扣（LeetCode）

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);
        vector<vector<int>> rel;
        if(intervals.size() == 1) {
            rel.push_back(intervals[0]);
            return rel;
        }
        for(int i = 1;i < intervals.size();i++) {
            if(intervals[i][0] <= intervals[i - 1][1]) {
                intervals[i][0] = intervals[i - 1][0];
                intervals[i][1] = max(intervals[i][1], intervals[i - 1][1]);
            }else {
                rel.push_back(intervals[i - 1]);
            }
        }
        rel.push_back(intervals[intervals.size() - 1]);
        return rel;
    }
};