【Leetcode】Letter Combinations of a Phone Number算法解析

用栈的思想。

第一种方法，设置两个string栈，每次从栈A取出string，加上一个char后放入栈B（注意放入栈B的是多个string，因为加char有超过一种加法例如数字1对应可以加'a'或'b'或'c'），直到栈A为空。栈A为空后把栈A和栈B swap一下。这样下一次循环还是从栈A去，往栈B放。

共循环n次，n是phone number的长度。

代码如下：

class Solution {public:    vector<string> letterCombinations(string digits) {        vector<string> stack1, stack2;                int max = digits.size();        stack1.push_back(string(""));        for(int i=0; i<max; i++) {            while(!stack1.empty()) {                string top = stack1.back();                stack1.pop_back();                                char digit = digits[i];                switch (digit) {                    case '2':                        stack2.push_back(top + 'a');                        stack2.push_back(top + 'b');                        stack2.push_back(top + 'c');                        break;                    case '3':                        stack2.push_back(top + 'd');                        stack2.push_back(top + 'e');                        stack2.push_back(top + 'f');                        break;                    case '4':                        stack2.push_back(top + 'g');                        stack2.push_back(top + 'h');                        stack2.push_back(top + 'i');                        break;                    case '5':                        stack2.push_back(top + 'j');                        stack2.push_back(top + 'k');                        stack2.push_back(top + 'l');                        break;                    case '6':                        stack2.push_back(top + 'm');                        stack2.push_back(top + 'n');                        stack2.push_back(top + 'o');                        break;                    case '7':                        stack2.push_back(top + 'p');                        stack2.push_back(top + 'q');                        stack2.push_back(top + 'r');                        stack2.push_back(top + 's');                        break;                    case '8':                        stack2.push_back(top + 't');                        stack2.push_back(top + 'u');                        stack2.push_back(top + 'v');                        break;                    case '9':                        stack2.push_back(top + 'w');                        stack2.push_back(top + 'x');                        stack2.push_back(top + 'y');                        stack2.push_back(top + 'z');                        break;                }             }            swap(stack1, stack2);        }        return stack1;    }};

第二种方法只用一个栈，即从这个栈取也往这个栈里放，取出来以后看这个string的长度，决定往里添加哪个digit对应的letter。例如输入数字是9893654，取出来一个string是sah，即长度为3，说明需要添加第4个digit对应的letter，即9893654中的3对应的letter。

但是要对输入的phone number进行一些处理——先在phone number里去掉0和1，因为这两个数没有对应任何字母。

代码如下：

class Solution {public:    vector<string> letterCombinations(string digits) {        vector<string> res;        vector<string> stack;                string::iterator pend = remove(digits.begin(), digits.end(), '0');        pend = remove(digits.begin(), pend, '1'); // 输入的digits中去除0和1        int max = pend - digits.begin();        stack.push_back(string(""));        while(!stack.empty()) {            string top = stack.back();            stack.pop_back();            int len = top.size();            if(len == max) {                res.push_back(top);                continue;            }                        char digit = digits[len];            switch (digit) {                case '2':                    stack.push_back(top + 'a');                    stack.push_back(top + 'b');                    stack.push_back(top + 'c');                    break;                case '3':                    stack.push_back(top + 'd');                    stack.push_back(top + 'e');                    stack.push_back(top + 'f');                    break;                case '4':                    stack.push_back(top + 'g');                    stack.push_back(top + 'h');                    stack.push_back(top + 'i');                    break;                case '5':                    stack.push_back(top + 'j');                    stack.push_back(top + 'k');                    stack.push_back(top + 'l');                    break;                case '6':                    stack.push_back(top + 'm');                    stack.push_back(top + 'n');                    stack.push_back(top + 'o');                    break;                case '7':                    stack.push_back(top + 'p');                    stack.push_back(top + 'q');                    stack.push_back(top + 'r');                    stack.push_back(top + 's');                    break;                case '8':                    stack.push_back(top + 't');                    stack.push_back(top + 'u');                    stack.push_back(top + 'v');                    break;                case '9':                    stack.push_back(top + 'w');                    stack.push_back(top + 'x');                    stack.push_back(top + 'y');                    stack.push_back(top + 'z');                    break;            }         }        return res;    }};