Code – 第95页

[note] 驻点不一定是极值点，极值点也不一定是驻点

1. 驻点不一定是极值点
因为不代表在左右变号，如在处是驻点但非极值点

2. 极值点也不一定是驻点
因为极值点有可能是不可导点，如在处是极值点但非驻点

LeetCode 553. Optimal Division

Given a list of positive integers, the adjacent integers will perform the float division. For example, [2,3,4] -> 2 / 3 / 4.

However, you can add any number of parenthesis at any position to change the priority of operations. You should find out how to add parenthesis to get the maximum result, and return the corresponding expression in string format. Your expression should NOT contain redundant parenthesis.

Example:
Input: [1000,100,10,2]
Output: “1000/(100/10/2)”
Explanation:
1000/(100/10/2) = 1000/((100/10)/2) = 200
However, the bold parenthesis in “1000/((100/10)/2)” are redundant,
since they don’t influence the operation priority. So you should return “1000/(100/10/2)”.

Other cases:
1000/(100/10)/2 = 50
1000/(100/(10/2)) = 50
1000/100/10/2 = 0.5
1000/100/(10/2) = 2
Note:

The length of the input array is [1, 10].
Elements in the given array will be in range [2, 1000].
There is only one optimal division for each test case.

题目大意：给定一个正整数列表，相邻的整数将执行浮点除法。例如[2,3,4] – > 2/3/4。但是，您可以在任何位置添加任意数量的括号以更改操作的优先级。您应该找到如何添加括号以获得最大结果，并以字符串格式返回相应的表达式，你的表达不应该包含多余的括号。

分析：要想得到最大的结果，只要使除数尽可能大，被除数尽可能小。被除过的数一定会变得更小，所以括号加在第一个数后面，括号内的数按从前到后顺序（不用添加冗余的括号）即可～

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

class Solution {

public:

string optimalDivision(vector<int>& nums) {

string ans = to_string(nums[0]);

if (nums.size() == 1) return ans;

if (nums.size() == 2) return ans + "/" + to_string(nums[1]);

ans = ans + "/(" + to_string(nums[1]);

for (int i = 2; i < nums.size(); i++) {

ans = ans + "/" + to_string(nums[i]);

}

return ans + ")";

}

};

LeetCode 647. Palindromic Substrings

Given a string, your task is to count how many palindromic substrings in this string.

The substrings with different start indexes or end indexes are counted as different substrings even they consist of same characters.

Example 1:
Input: “abc”
Output: 3
Explanation: Three palindromic strings: “a”, “b”, “c”.
Example 2:
Input: “aaa”
Output: 6
Explanation: Six palindromic strings: “a”, “a”, “a”, “aa”, “aa”, “aaa”.
Note:
The input string length won’t exceed 1000.

题目大意：给定一个字符串，计算这个字符串中有多少回文子串。（具有不同start下标或end下标的子字符串即使包含相同的字符也会被计为不同的子字符串。）
分析：对于s[i]来说，检测s[i-j]==s[i+j]是否成立，每一次相等就ans++、检测s[i-j-1]==s[i+j]是否成立，每次成立就ans++，最后返回ans的值～

class Solution {
public:
    int countSubstrings(string s) {
        int ans = 0, n = s.length();
        for (int i = 0; i < n; i++) {
            for (int j = 0; i + j < n && i - j >= 0 && s[i-j] == s[i+j]; j++) ans++;
            for (int j = 0; i + j < n && i - j - 1 >=0 && s[i-j-1] == s[i+j]; j++) ans++;
        }
        return ans;
    }
};

class Solution {

public:

int countSubstrings(string s) {

int ans = 0, n = s.length();

for (int i = 0; i < n; i++) {

for (int j = 0; i + j < n && i - j >= 0 && s[i-j] == s[i+j]; j++) ans++;

for (int j = 0; i + j < n && i - j - 1 >=0 && s[i-j-1] == s[i+j]; j++) ans++;

}

return ans;

}

};

LeetCode 547. Friend Circles

There are N students in a class. Some of them are friends, while some are not. Their friendship is transitive in nature. For example, if A is a direct friend of B, and B is a direct friend of C, then A is an indirect friend of C. And we defined a friend circle is a group of students who are direct or indirect friends.

Given a N*N matrix M representing the friend relationship between students in the class. If M[i][j] = 1, then the ith and jth students are direct friends with each other, otherwise not. And you have to output the total number of friend circles among all the students.

Example 1:
Input:
[[1,1,0],
[1,1,0],
[0,0,1]]
Output: 2
Explanation:The 0th and 1st students are direct friends, so they are in a friend circle.
The 2nd student himself is in a friend circle. So return 2.
Example 2:
Input:
[[1,1,0],
[1,1,1],
[0,1,1]]
Output: 1
Explanation:The 0th and 1st students are direct friends, the 1st and 2nd students are direct friends,
so the 0th and 2nd students are indirect friends. All of them are in the same friend circle, so return 1.
Note:
N is in range [1,200].
M[i][i] = 1 for all students.
If M[i][j] = 1, then M[j][i] = 1.

题目大意：班上有N个学生。有些是朋友，有些则不是。他们的友谊本质上是传递性的。例如，如果A是B的直接朋友，B是C的直接朋友，那么A是C的间接朋友。我们定义了一个朋友圈是一群直接或间接的朋友。给定N * N矩阵M代表班级中学生之间的朋友关系。如果M[i][j] = 1，那么第i和第j个学生是彼此直接的朋友，否则不是，输出所有学生中的朋友圈的总数

分析：用并查集，cnt一开始为n，每个人的父亲也都是自己，将每一对朋友的父亲结点找到，如果他们的父亲结点不是同一个，那么就合并为一个集合，并将cnt-1，最后输出cnt

class Solution {
public:
    int findCircleNum(vector<vector<int>>& M) {
        int n = M.size(), cnt = M.size();
        father.resize(n);
        for (int i = 0; i < n; i++) father[i] = i;
        for (int i = 0; i < n; i++) {
            for (int j = i + 1; j < n; j++) {
                if (M[i][j] == 1) {
                    int faA = findFather(i);
                    int faB = findFather(j);
                    if (faA != faB) {
                        father[faA] = father[faB];
                        cnt--;
                    }
                }
                
            }
        }
        return cnt;
    }
private:
    vector<int> father;
    int findFather(int x) {
        return x == father[x] ? x : findFather(father[x]);
    }
};

class Solution {

public:

int findCircleNum(vector<vector<int>>& M) {

int n = M.size(), cnt = M.size();

father.resize(n);

for (int i = 0; i < n; i++) father[i] = i;

for (int i = 0; i < n; i++) {

for (int j = i + 1; j < n; j++) {

if (M[i][j] == 1) {

int faA = findFather(i);

int faB = findFather(j);

if (faA != faB) {

father[faA] = father[faB];

cnt--;

}

return cnt;

}

private:

vector<int> father;

int findFather(int x) {

return x == father[x] ? x : findFather(father[x]);

}

};

LeetCode 240. Search a 2D Matrix II

Write an efficient algorithm that searches for a value in an m x n matrix. This matrix has the following properties:

Integers in each row are sorted in ascending from left to right.
Integers in each column are sorted in ascending from top to bottom.
For example,

Consider the following matrix:

[
[1, 4, 7, 11, 15],
[2, 5, 8, 12, 19],
[3, 6, 9, 16, 22],
[10, 13, 14, 17, 24],
[18, 21, 23, 26, 30]
]
Given target = 5, return true.

Given target = 20, return false.

题目大意：编写一个有效的算法，在m×n矩阵中搜索一个值。该矩阵具有以下属性：
每行中的整数从左到右依次排列。每列中的整数按从上到下的顺序排列。

分析：对每一行进行binary_serch()，如果能够找到则return true，否则返回false

class Solution {
public:
    bool searchMatrix(vector<vector<int>>& matrix, int target) {
        for(int i = 0; i < matrix.size(); i++)
            if (binary_search(matrix[i].begin(), matrix[i].end(), target)) return true;
        return false;
    }
};

class Solution {

public:

bool searchMatrix(vector<vector<int>>& matrix, int target) {

for(int i = 0; i < matrix.size(); i++)

if (binary_search(matrix[i].begin(), matrix[i].end(), target)) return true;

return false;

}

};

LeetCode 74. Search a 2D Matrix

Write an efficient algorithm that searches for a value in an m x n matrix. This matrix has the following properties:

Integers in each row are sorted from left to right.
The first integer of each row is greater than the last integer of the previous row.
For example,

Consider the following matrix:

[
[1, 3, 5, 7],
[10, 11, 16, 20],
[23, 30, 34, 50]
]
Given target = 3, return true.

题目大意：编写一个有效的算法，在m×n矩阵中搜索一个值。该矩阵具有以下属性：
每行中的整数从左到右排序。每行的第一个整数大于前一行的最后一个整数。

分析：对每一行进行二分搜索，如果找到了就返回true否则返回false～

class Solution {
public:
    bool searchMatrix(vector<vector<int>>& matrix, int target) {
        for(int i = 0; i < matrix.size(); i++)
            if (binary_search(matrix[i].begin(), matrix[i].end(), target)) return true;
        return false;
    }
};

class Solution {

public:

bool searchMatrix(vector<vector<int>>& matrix, int target) {

for(int i = 0; i < matrix.size(); i++)

if (binary_search(matrix[i].begin(), matrix[i].end(), target)) return true;

return false;

}

};