第 93 场双周赛

1.数组中字符串的最大值

数组中字符串的最大值

Solution

class Solution {
public:
    int maximumValue(vector<string>& strs) {
        int ans = -1;
        for(auto x : strs){
            int n = x.size();
            int sum = 0;
            bool flag = true;
            for(int i = 0;i < n;i++){
                if(x[i] < '0' || x[i] > '9'){
                    ans = max(ans,n);
                    flag = false;
                    break;
                }
                sum = sum*10 + x[i]-'0';
            }
            if(flag) ans = max(ans,sum);
        }
        return ans;
    }
};

2.图中最大星和

图中最大星和

Solution

建立图的邻接表，然后按大小排序算就行

class Solution {
public:
    int maxStarSum(vector<int>& vals, vector<vector<int>>& edges, int k) {
        //建立邻接表
        int n = vals.size();
        vector<vector<int>> g(n);
        for (auto &e : edges) {
            if (vals[e[1]] > 0) {
                g[e[0]].push_back(vals[e[1]]);
            }
            if (vals[e[0]] > 0) {
                g[e[1]].push_back(vals[e[0]]);
            }
        }
        int ans = INT_MIN;
        for (int i = 0; i < n; i++) {
            int cur = vals[i];
            sort(g[i].begin(), g[i].end(), greater<>());
            int upper = min(k, (int)g[i].size());
            for (int j = 0; j < upper; j++) {
                cur += g[i][j];
            }
            ans = max(ans, cur);
        }
        return ans;
    }
};

3.青蛙过河 II

青蛙过河 II

Solution

二分答案，看题就能看出

class Solution {
public:
    int maxJump(vector<int>& stones) {
        auto check = [&](int x, vector<int> &m){
            //先跳到最后
            vector<bool> book(m.size(),false);
            for(int i = 0,j = 1;j < m.size();j++){
                while(j < m.size() && m[j] - m[i] <= x) j++;
                book[j-1] = true;
                i = j-1;
            }
            book[m.size() - 1] = false;
            //从正面跳一遍
            for(int i = 0,j = 1;j < m.size(); j++){
                while(j < m.size() && book[j]) j++;
                if(m[j] - m[i] > x) return false;
                i = j;
            }
            return true;
        };
        int l = 1, r = stones.back();
        while(l < r){
            int mid = l + (r - l) / 2;
            if(check(mid,stones)){
                r = mid;
            }else{
                l = mid+1;
            }
        }
        return l;
    }
};

4.让数组不相等的最小总代价

让数组不相等的最小总代价

Solution

class Solution {
public:
    long long minimumTotalCost(vector<int>& nums1, vector<int>& nums2) {
        long long num = nums1[0],cnt = 0;
        vector<int> point;
        //跑出众数
        for(int i = 0;i < nums1.size();i++){
            if(nums1[i] == nums2[i]){
                point.push_back(i);
                if(cnt == 0) num = nums1[i];
                if(num == nums1[i]) cnt++;
                else cnt--;
            }
        }
        cnt = 0;
        //记录众数数量
        for(int i = 0;i < nums1.size();i++){
            if(nums1[i] == nums2[i] && num == nums1[i]){
                cnt++;
            }
        }
        int n = point.size();
        long long ans = 0;
        if(cnt >= ceil(n / 2.0)){//超过一半需要从左到右进行选择
            ans = accumulate(point.begin(),point.end(),0ll);
            cnt = cnt - (n - cnt);
            for(int i = 0;i < nums1.size();i++){
                if(nums1[i] != nums2[i] && num != nums1[i] && num != nums2[i]){
                    cnt--;ans += i;
                }
                if(cnt == 0) break;
            }
            if(cnt > 0)return -1;
        }
        else{
            ans = accumulate(point.begin(),point.end(),0ll);
        }
        return ans;
    }
};