CCI_chapter 1

1.1Implement an algorithm to determine if a string has all unique characters What if you can not use additional data structures?

bool isUniqueChars(string str) {
        unsigned int checklittle = 0;
        unsigned int checklarger = 0;
        for(int i = 0; i < str.size();i++)
        {   
            bool flag  = str[i] - 'a' >= 0 ;
            unsigned int temp;
            temp = flag ?  1 << (str[i] - 'a') :
                            1 << (str[i] - 'A');
            if(flag){
                if( checklittle & temp ) return false;
                else  checklittle |= temp;
            }else {

                if(checklarger & temp) return false;
                else checklarger |= temp;
            }
        }
        return true;

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1.2 Write code to reverse a C-Style String

void reverse(char *str){
        if(NULL == str) return;
        char *p = str;
        while(*p)++p;
        --p;
        while(str < p){
            char temp = *p;
            *p = *str;
            *str = temp;
            --p;
            ++str;
        }
    }

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1.3 Design an algorithm and write code to remove the duplicate characters in a string without using any additional bufer NOTE: One or two additional variables are fine .An extra copy of the array is not

void removeDuplicates(char[] str){
    if(NULL == str) return ;
    int len  = strlen(str);
    if(len < 2) return ;
    int tail = 1;
    for(int i = 1; i < len ; i++){
        int j;
        for(j = 0; j < tail ; j++){
            if(str[j] == str[i]) break;
        }
        if(j == tail){
            str[tail] = str[i];
            ++tail;
        }
    }
    str[tail] = '';
}

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这道题里面判断重复也可以使用1.1的思想，不过要提前搞清楚输入参数的字符集

1.4Write a method to decide if two strings are anagrams or not

bool anagram(string s, string t){

    if( s.size() != t.size()) return false;
    if( s.size() == 0) return true;

    int  table[256];
    memset(table, 0, sizeof(int) * 256);
    for(char c : s){
        table[c]++;
    }
    for(char c: t){
        table[c]--;
    }
    for (int c : table){
        if(c != 0) return false;
    }

    return true;
}

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1.5

1.6Given an image represented by an NxN matrix, where each pixel in the image is 4 bytes, write a method to rotate the image by 90 degrees Can you do this in place?

void rotate(vector<vector<int> > &matrix) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        int n = matrix.size();
        if(n <= 1) return;

        for( int lay = 0; lay < n/2; lay++){ 
            int start = lay;
            int end = n - lay -1;
            for(int i = start ; i < end ; i++){
                // record top
                int temp = matrix[start][i];
                //left to top
                matrix[start][i] = matrix[n -1- i][start];
                // bottom to left
                matrix[n -1- i][start] = matrix[end][n -1-i];
                //right to bottom
                matrix[end][n-1-i] = matrix[i][end];
                // top to right
                matrix[i][end] = temp;
            }
        }
    }

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1.7Write an algorithm such that if an element in an MxN matrix is 0, its entire row and column is set to 0

void setZeroes(vector<vector<int> > &matrix) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        int m = matrix.size();
        if(m < 1) return ;
        int n = matrix[0].size();

        bool zeroR = false, zeroC = false;
        for(int i = 0; i< n ; i++)
                if(matrix[0][i] == 0){
                    zeroR = true;
                    break;
                }
        for( int i = 0; i< m ; i++){
            if(matrix[i][0] ==0){
                    zeroC = true;
                    break;
            }
        }
        for(int i = 1; i< m; i++)
          for(int j = 1; j< n; j++)
                if(matrix[i][j] == 0){
                    matrix[0][j] = 0;
                    matrix[i][0] = 0;
                }
        for(int i = 1; i< m;i++)
            for( int j = 1; j< n; j++)
                if(matrix[0][j] ==0 || matrix[i][0] ==0 )
                    matrix[i][j] = 0;

        for(int i = 0; i< n && zeroR ; i++) matrix[0][i] = 0;
        for(int i = 0; i< m && zeroC ; i++) matrix[i][0] = 0;
    }

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1.8 Assume you have a method isSubstring which checks if one word is a substring of another Given two strings, s1 and s2, write code to check if s2 is a rotation of s1 using

only one call to isSubstring (i e , “waterbottle” is a rotation of “erbottlewat”)

bool isRotation(string s1, string s2){
        if(s1.size() != s2.size()) return false;
        if(s1.size() == 0) return true;
        string ss = s1+ s1;
        if(string::npos != ss.find(s2)) return true;
        return false ;
}

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原文链接: https://www.cnblogs.com/graph/p/3264330.html

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