Leetcode第 296 场周赛

class Solution {
public:
    int minMaxGame(vector& nums) {
        if(nums.size()==1) return nums[0];
        vector v;
        for(int i=0;i
            if((i/2)%2==0) v.push_back(min(nums[i],nums[i+1]));
            else v.push_back(max(nums[i],nums[i+1]));
        }
        return minMaxGame(v);
    }
};

class Solution {
public:
    int partitionArray(vector& nums, int k) {
        sort(nums.begin(),nums.end());
        int ans = 1;
        int left = nums[0];
        for(int i=0;i
            if(nums[i]-left>k) {
                ans++;
                left = nums[i];
            }
        }
        return ans;
    }
};

可以map写是：题目保证在第 i 个操作中：
operations[i][0] 在 nums 中存在。
operations[i][1] 在 nums 中不存在。

class Solution {
public:
    vector arrayChange(vector& nums, vector>& operations) {
        unordered_map mp;
        for(int i=0;i
            nums[mp[operations[i][0]]] = operations[i][1];
            mp[operations[i][1]] = mp[operations[i][0]];
            mp.erase(operations[i][0]);            
        }
        return nums;
    }
};

比赛的时候想着这个思路，然后用substr一直超时 后来看榜单代码 直接k操作 回头一看k<=40. 后来看超时的样例 字符很长 我调用substr(0,ix) 这是线性操作难免超时

class TextEditor {
public:
    string left,right;
    int ix;
    TextEditor() {
        left = right = "";
    }
    void addText(string text) {
        left += text;
    }
    int deleteText(int k) {
        int len = left.size();
        k = min(k,len);
        for(int i=0;i
        int len = left.size();
        k = min(k,len);
        for(int i=0;i
            right.push_back(left.back());
            left.pop_back();
        }
        string ans(left.end()-min(10,int(left.size())),left.end());
        return ans;
    }
    string cursorRight(int k) {
        int len = right.size();
        k = min(k,len);
        for(int i=0;i
            left.push_back(right.back());
            right.pop_back();
        }
        string ans(left.end()-min(10,int(left.size())),left.end());
        return ans;
    }
};