1. 題目

求解一個給定的方程，將x以字符串"x=#value"的形式返回。該方程僅包含’+’，’ - '操作，變量 x 和其對應系數。

如果方程沒有解，請返回“No solution”。

如果方程有無限解，則返回“Infinite solutions”。

如果方程中只有一個解，保證返回值 x 是一個整數。

示例 1： 輸入: "x+5-3+x=6+x-2" 輸出: "x=2"示例 2: 輸入: "x=x" 輸出: "Infinite solutions"示例 3: 輸入: "2x=x" 輸出: "x=0"示例 4: 輸入: "2x+3x-6x=x+2" 輸出: "x=-1"示例 5: 輸入: "x=x+2" 輸出: "No solution"

來源：力扣（LeetCode） 鏈接：https://leetcode-cn.com/problems/solve-the-equation
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2. 解題

• 寫的不是很簡潔，可優化
• 按=號分成兩邊，把系數加總

class Solution { public:string solveEquation(string equation) {int i, lnum = 0, rnum = 0, lcoe = 0, rcoe = 0, n=0;char ch;bool positive = true;for(i = 0; equation[i] != '='; ++i){ch = equation[i];if(ch == 'x'){lcoe += n==0 ? ((i>0&&equation[i-1]=='0')? 0 : (positive ? 1 : -1)) : (positive ? n : -n);// 注意考慮 "0x=0" n=0;}else if(ch == '-'){lnum += (positive ? n : -n);positive = false;n=0;}else if(ch == '+'){lnum += (positive ? n : -n);positive = true;n=0;}elsen = 10*n+ch-'0';}if(equation[i-1] != 'x')lnum += (positive ? n : -n);positive = true;n = 0;for(i++; i < equation.size(); ++i){ch = equation[i];if(ch == 'x'){rcoe += n==0 ? ((equation[i-1]=='0')? 0 : (positive ? 1 : -1)) : (positive ? n : -n);n=0;}else if(ch == '-'){rnum += (positive ? n : -n);positive = false;n=0;}else if(ch == '+'){rnum += (positive ? n : -n);positive = true;n=0;}elsen = 10*n+ch-'0';}if(equation[i-1] != 'x')rnum += (positive ? n : -n);if(lcoe == rcoe && lnum == rnum)return "Infinite solutions";if(lcoe == rcoe && lnum != rnum)return "No solution";int ans = (lnum-rnum)/(rcoe-lcoe);return "x="+to_string(ans);} };

0 ms 6.1 MB

• 優化代碼重復段

class Solution { public:string solveEquation(string equation) {int i, lnum = 0, rnum = 0, lcoe = 0, rcoe = 0, n=0;int pos = equation.find('=');vector<int> coeff = cal_coeff(equation.substr(0,pos));lcoe = coeff[0];lnum = coeff[1];coeff = cal_coeff(equation.substr(pos+1));rcoe = coeff[0];rnum = coeff[1];if(lcoe == rcoe && lnum == rnum)return "Infinite solutions";if(lcoe == rcoe && lnum != rnum)return "No solution";int ans = (lnum-rnum)/(rcoe-lcoe);return "x="+to_string(ans);}vector<int> cal_coeff(string equation){char ch;bool positive = true;int i, n = 0, coe = 0, num = 0;for(i = 0; i < equation.size(); ++i){ch = equation[i];if(ch == 'x'){coe += n==0 ? ((i>0&&equation[i-1]=='0')? 0 : (positive ? 1 : -1)) : (positive ? n : -n);n=0;}else if(ch == '-'){num += (positive ? n : -n);positive = false;n=0;}else if(ch == '+'){num += (positive ? n : -n);positive = true;n=0;}elsen = 10*n+ch-'0';}if(equation[i-1] != 'x')num += (positive ? n : -n);return {coe, num};} };

總結

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