455. 分发饼干

饼干从大的开始利用,优先满足胃口大的;

class Solution {
public:
    int findContentChildren(vector<int>& g, vector<int>& s) {
        sort(g.begin(),g.end());
        sort(s.begin(),s.end());
        int res=0;
        int index=s.size()-1;
        for(int i=g.size()-1;i>=0;i--){
            if(index>=0&&s[index]>=g[i]) {//注意index不要越界,index的范围
                res+=1;
                index-=1;
            }
        }
        return res;
    }
};

另一种方法:

饼干从小的开始利用,优先满足胃口小的;

class Solution {
public:
    int findContentChildren(vector<int>& g, vector<int>& s) {
        sort(g.begin(),g.end());
        sort(s.begin(),s.end());
        int index=0;
        int res=0;
        for(int i=0;i<s.size();i++){
            if(index<g.size()&&g[index]<=s[i]){
                res+=1;
                index+=1;
            }
        }
        return res;
    }
};

376. 摆动序列

class Solution {
public:
    int wiggleMaxLength(vector<int>& nums) {
        int res=1;
        if(nums.size()<=1) return nums.size();
        int pre=0;
        int cur=0;
        for(int i=0;i<nums.size()-1;i++){
            cur=nums[i+1]-nums[i];
            if((pre>=0&&cur<0)||(pre<=0&&cur>0)){
                res++;
                pre=cur;
            }
        }
        return res;
    }
};

53. 最大子数组和

class Solution {
public:
    int maxSubArray(vector<int>& nums) {
        int res=INT_MIN;
        int count=0;
        for(int i=0;i<nums.size();i++){
            count+=nums[i];
            if(count>res)res=count;
            if(count<0) count=0;
        }
        return res;
    }
};

122. 买卖股票的最佳时机 II

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

55. 跳跃游戏

class Solution {
public:
    bool canJump(vector<int>& nums) {
        int cover=0;
        for(int i=0;i<=cover;i++){
            cover=max(i+nums[i],cover);
            if(cover>=nums.size()-1) return true;
        }
        return false;
    }
};

45. 跳跃游戏 II

class Solution {
public:
    int jump(vector<int>& nums) {
        int res=0;
        int cur=0;
        int next=0;
        if(nums.size()==1)return 0;
        for(int i=0;i<nums.size();i++){
            next=max(i+nums[i],next);
            if(i==cur){
                res++;
                cur=next;
                if(cur>=nums.size()-1){
                    break;
                }
            }
        }
        return res;
    }
};

1005. K 次取反后最大化的数组和

class Solution {
public:
    static bool cmp(int a,int b){
        return abs(a)>abs(b);
    }
    int largestSumAfterKNegations(vector<int>& nums, int k) {
        sort(nums.begin(),nums.end(),cmp);
        for(int i=0;i<nums.size();i++){
            if(nums[i]<0&&k>0){
                nums[i]*=-1;
                k--;
            }
        }
        if(k%2==1){
            nums[nums.size()-1]*=-1;
        }
        int res=0;
        for(int i=0;i<nums.size();i++){
            res+=nums[i];
        }
        return res;
    }
};

134. 加油站

感觉加油站这题做的很晕

class Solution {
public:
    //1.从0遍历结束之后,rest>=0,就返回0
    //2.从0遍历结束之后,rest<0,返回-1
    //3.从0遍历,发现中间有rest<0,但整体>=0,那就从后遍历找rest累计能补掉这个漏洞的
    int canCompleteCircuit(vector<int>& gas, vector<int>& cost) {
        int curSum=0;
        int min=INT_MAX;
        for(int i=0;i<cost.size();i++){
            int rest=gas[i]-cost[i];
            curSum+=rest;
            if(curSum<min){
                min=curSum;
            }
        }
        if(min>=0)return 0;
        if(curSum<0)return -1;
        for(int i=cost.size()-1;i>=0;i--){
            int rest=gas[i]-cost[i];
            min+=rest;
            if(min>=0)return i;
        }
        return -1;
    }
};

方法2:

class Solution {
public:
    int canCompleteCircuit(vector<int>& gas, vector<int>& cost) {
        int curSum=0;
        int totalSum=0;
        int start=0;
        for(int i=0;i<gas.size();i++){
            curSum+=gas[i]-cost[i];
            totalSum+=gas[i]-cost[i];
            if(curSum<0){
                start=i+1;
                curSum=0;
            }
        }
        if(totalSum<0) return -1;
        return start;
    }
};

135. 分发糖果

class Solution {
public:
    int candy(vector<int>& ratings) {
        vector<int>candyVec(ratings.size(),1);
        for(int i=1;i<ratings.size();i++){
            if(ratings[i]>ratings[i-1]) candyVec[i]=candyVec[i-1]+1;
        }
        //从后往前遍历
        for(int i=ratings.size()-2;i>=0;i--){
            if(ratings[i]>ratings[i+1]) candyVec[i]=max(candyVec[i],candyVec[i+1]+1);
        }
        int res=0;
        for(int i=0;i<ratings.size();i++){
            res+=candyVec[i];
        }
        return res;
    }
};

860. 柠檬水找零

class Solution {
public:
    bool lemonadeChange(vector<int>& bills) {
        int five=0,ten=0,twenty=0;
        for(int i=0;i<bills.size();i++){
            if(bills[i]==5) five++;
            
            if(bills[i]==10){
                if(five==0)return false;
                else {
                    five--;
                    ten++;
                }
            }

            if(bills[i]==20){
                if(ten>0&&five>0){
                    ten--;
                    five--;
                    twenty++;
                }
                else if(five>=3){
                    five-=3;
                    twenty++;
                }
                else return false;
            }
        }
        return true;
    }
};

406. 根据身高重建队列

class Solution {
public:
    static bool cmp(const vector<int>& a,const vector<int>& b){
        if(a[0]==b[0])return a[1]<b[1];
        return a[0]>b[0];
    }
    vector<vector<int>> reconstructQueue(vector<vector<int>>& people) {
        sort(people.begin(),people.end(),cmp);
        vector<vector<int>>que;
        for(int i=0;i<people.size();i++){
            int position=people[i][1];
            que.insert(que.begin()+position,people[i]);
        }
        return que;
    }
};

改进:容器使用了list,list底层是链表实现的,比起vector能减少时间复杂度

class Solution {
public:
    static bool cmp(const vector<int>& a,const vector<int>& b){
        if(a[0]==b[0])return a[1]<b[1];
        return a[0]>b[0];
    }
    vector<vector<int>> reconstructQueue(vector<vector<int>>& people) {
        sort(people.begin(),people.end(),cmp);
        list<vector<int>>que;
        for(int i=0;i<people.size();i++){
            int position=people[i][1];
            auto it=que.begin();
            while(position--){
                it++;
            }
            que.insert(it,people[i]);
        }
        return vector<vector<int>>(que.begin(),que.end());
    }
};

452. 用最少数量的箭引爆气球

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

435. 无重叠区间

按照区间末端进行排序

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

按照区间前段进行排序:

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

763. 划分字母区间

class Solution {
public:
    vector<int> partitionLabels(string s) {
        //初始化每个字符的最远位置
        int hash[26]={0};
        //遍历 初始化
        for(int i=0;i<s.size();i++){
            hash[s[i]-'a']=i;
        }
        //遍历,寻找最远距离
        vector<int>res;
        int left=0;
        int right=0;

        for(int i=0;i<s.size();i++){
            right=max(right,hash[s[i]-'a']);
            if(right==i){
                res.push_back(right-left+1);
                left=right+1;
            }
            
        }
        return res;
    }
};

56. 合并区间

class Solution {
public:
    static bool cmp(const vector<int>& a,const 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>>res;
        res.push_back(intervals[0]);
        for(int i=1;i<intervals.size();i++){
            if(intervals[i][0]<=res.back()[1]){
                res.back()[1]=max(res.back()[1],intervals[i][1]);
            }
            else {
                res.push_back(intervals[i]);
            }
        }
        return res;
    }
};

738. 单调递增的数字

class Solution {
public:
    int monotoneIncreasingDigits(int n) {
        string strNum=to_string(n);
        //flag标记从哪里开始被赋值成9
        int flag=strNum.size();
        for(int i=strNum.size()-1;i>0;i--){
            if(strNum[i]<strNum[i-1]){
                flag=i;
                strNum[i-1]--;
            }
        }

        for(int i=flag;i<strNum.size();i++){
            strNum[i]='9';
        }
        return stoi(strNum);
    }
};

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