1.最长回文串

最长回文串

class Solution {
public:int longestPalindrome(string s) {//计数一：用数组模拟哈希表int hash[127] = {0};for(auto x:s){hash[x]++;}//统计结果int ret = 0;for(auto x:hash){ret += x/2*2;}return ret<s.size()?ret+1:ret;}
};

2.增减字符串匹配

增减字符串匹配

class Solution {
public:vector<int> diStringMatch(string s) {//贪心//遇到I，选择当前能选择的最小的数//遇到D，选择当前能选择的最大的数int left = 0,right = s.size();vector<int> ret;for(auto ch:s){if(ch == 'I') ret.push_back(left++);else ret.push_back(right--);}ret.push_back(left);return ret;}
};

3.分发饼干

分发饼干

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

4.最优除法

最优除法

class Solution {
public:string optimalDivision(vector<int>& nums) {//贪心+找规律int n = nums.size();if(n == 1) return to_string(nums[0]);if(n == 2) return to_string(nums[0])+'/'+to_string(nums[1]);string str = to_string(nums[0])+"/("+to_string(nums[1]);for(int i=2;i<n;i++){str+='/'+to_string(nums[i]);}str +=')';return str;}
};

5.跳跃游戏II

跳跃游戏II

class Solution {
public:int jump(vector<int>& nums) {//使用层序遍历的思想，一层一层往后跳//maxPos表示下一层最右端点的下标int left = 0,right = 0,maxPos = 0,ret = 0,n = nums.size();while(left<=right){if(maxPos>=n-1) return ret;//判断能否跳到最后一个位置//遍历当前层for(int i=left;i<=right;i++){maxPos = max(maxPos,nums[i]+i);}left = right+1;right = maxPos;ret++;}return -1;}
};

6.跳跃游戏

跳跃游戏

class Solution {
public:bool canJump(vector<int>& nums) {//跟上一题跳跃游戏II思路一模一样//一层一层往后跳，层序遍历int left = 0,right = 0,maxPos = 0,n =nums.size();while(left<=right){if(maxPos>=n-1) return true;for(int i=left;i<=right;i++){maxPos = max(maxPos,nums[i]+i);}left = right+1;right = maxPos;}return false;}
};

7.加油站

加油站

class Solution {
public:int canCompleteCircuit(vector<int>& gas, vector<int>& cost) {//解法一：暴力枚举 O(n^2) 会超时int n = gas.size();int step = n;for(int i=0;i<n;i++){int rest = 0;for(int step=0;step<n;step++){int index = (i+step)%n;rest = rest+gas[index]-cost[index];if(rest<0) break;}if(rest>=0) return i;}return -1;}
};

class Solution {
public:int canCompleteCircuit(vector<int>& gas, vector<int>& cost) {//解法二：找规律+在解法一的基础上稍作改动int n = gas.size();for(int i=0;i<n;i++){int rest = 0;int step = 0;for(;step<n;step++){int index = (i+step)%n;rest = rest+gas[index]-cost[index];if(rest<0) break;}if(rest>=0) return i;i = i+step;}return -1;}
};

8.单调递增的数字

单调递增的数字

class Solution {
public:int monotoneIncreasingDigits(int n) {//找规律string s = to_string(n);int i=0,m = s.size();//找到第一个递减的位置while(i+1<m && s[i]<=s[i+1]) i++;if(i+1 == m) return n;//回推while(i-1>=0 && s[i]==s[i-1]) i--;s[i]--;for(int j=i+1;j<m;j++){s[j]='9';}return stoi(s);}
};

9.坏了的计算器

坏了的计算器

class Solution {
public:int brokenCalc(int startValue, int target) {//正难则反+贪心//从end->begin *->/ -1->+1int ret = 0;while(target>startValue){if(target%2 == 0) target/=2;else target += 1;ret++;}return ret+startValue-target;}
};