标签：near for logs bst ++ tco etc exactly 计算
Solve a given equation and return the value of x in the form of string "x=#value". The equation contains only ‘+‘, ‘-‘ operation, the variable x and its coefficient.
If there is no solution for the equation, return "No solution".
If there are infinite solutions for the equation, return "Infinite solutions".
If there is exactly one solution for the equation, we ensure that the value of x is an integer.
Example 1:
Input: "x+5-3+x=6+x-2" Output: "x=2"
Example 2:
Input: "x=x" Output: "Infinite solutions"
Example 3:
Input: "2x=x" Output: "x=0"
Example 4:
Input: "2x+3x-6x=x+2" Output: "x=-1"
Example 5:
Input: "x=x+2" Output: "No solution"
模拟方程式的求解。需要对符号，字母，数字进行判断后统计。引入两个变量，一个记录x的系数，一个记录数字的总和。维护一个符号变量用于计算数字的总和，在等号左边符号变量为正，在等号右边符号变量为负。
class Solution { public: string solveEquation(string equation) { int n = equation.size(), sign = 1, coeff = 0, sum = 0, i = 0; for (int j = 0; j < n; j++) { // cal the sum of equation. if (equation[j] == ‘+‘ || equation[j] == ‘-‘) { if (j > i) sum += sign * stoi(equation.substr(i, j-i)); i = j; } // cal the coeff of equation. else if (equation[j] == ‘x‘) { if (i == j || equation[j - 1] == ‘+‘) coeff += sign; else if (equation[j - 1] == ‘-‘) coeff -= sign; else coeff += sign * stoi(equation.substr(i, j - i)); i = j + 1; } // flip the sign after meeting ‘=‘. else if (equation[j] == ‘=‘) { // calc the sum if the num near the ‘=‘ if (j > i) sum += sign * stoi(equation.substr(i, j - i)); sign = -1; i = j + 1; } } // cal the last num of equation. if (i < n) sum += sign * stoi(equation.substr(i)); // judge the coeff and sum. if (coeff == 0 && sum == 0) return "Infinite solutions"; if (coeff == 0 && sum != 0) return "No solution"; int res = -sum/coeff; return "x=" + to_string(res); } }; // 3 ms
标签：near for logs bst ++ tco etc exactly 计算
原文地址：http://www.cnblogs.com/immjc/p/7158013.html
