力扣hot100——贪心

121. 买卖股票的最佳时机

class Solution {
public:
    int maxProfit(vector<int>& a) {
        if (a.size() == 1) return 0;
        int ans = 0;
        int mi = a[0];
        for (int i = 1; i < a.size(); i++) {
            ans = max(ans, a[i] - mi);
            mi = min(mi, a[i]);
        }
        return ans;
    }
};

55. 跳跃游戏

class Solution {
public:
    bool canJump(vector<int>& a) {
        int r = 0;
        for (int i = 0; i < a.size(); i++) {
            if (r < i) return false;
            r = max(r, i + a[i]);
        }
        return true;
    }
};

45. 跳跃游戏 II

class Solution {
public:
    int jump(vector<int>& a) {
        int r = 0;
        int mx = 0;
        int ans = 0;
        for (int i = 0; i < a.size() - 1; i++) {
            mx = max(mx, i + a[i]);
            if (i == r) {
                ans++;
                r = mx;
            }
        }
        return ans;
    }
};

贪心， 求最小的覆盖区间，以至于覆盖所有区间

注意题目保证了一定可以到达，并且考虑边界情况，只需到到size - 2位置就行

 763. 划分字母区间

class Solution {
public:
    vector<int> partitionLabels(string s) {
        map<char, pair<int, int>> mp;
        for (int i = 0; i < s.size(); i++) {
            if (!mp.count(s[i])) mp[s[i]] = { i, i };
            else mp[s[i]].second = i;
        }
        vector<pair<int, int>> v;
        for (auto [x, y] : mp) v.push_back(y);
        sort(v.begin(), v.end());
        int l = 0;
        int r = 0;
        vector<int> ans;
        for (int i = 0; i < v.size(); i++) {
            if (v[i].first > r) {
                ans.push_back(r - l + 1);
                l = v[i].first;
            }
            r = max(r, v[i].second);
        }
        ans.push_back(r - l + 1);
        return ans;
    }
};

区间合并典题：求出同种字符的最大覆盖区间，合并区间即可