问题：

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"

解决：

① 解方程。难点在于处理字符串，如何将x的系数合并起来，将常数合并起来，化简成ax=b的形式来求解。

例：
x + 5 -2x = 6-3x;
leftPart：tokens= {x，+ 5，-2x}; 系数为x = 1-2 = -1; 常数= 5;
rightPart：tokens = {6，-3x}; 系数为x = -3; 常数= 6;

最终结果=（6-5）/（-1 - （ - 3））

class Solution { //12ms
    public String solveEquation(String equation) {
        String[] parts = equation.split("=");
        int[] leftPart = evaluate(parts[0]);
        int[] rightPart = evaluate(parts[1]);
        if (leftPart[0] == rightPart[0] && leftPart[1] == rightPart[1]){
            return "Infinite solutions";
        }else if (leftPart[0] == rightPart[0]){
            return "No solution";
        }
        return "x=" + (rightPart[1] - leftPart[1]) / (leftPart[0] - rightPart[0]);
    }
    public int[] evaluate(String str){
        String[] tokens = str.split("(?=[+-])");//（）匹配组; ？=匹配并包含在res中; [+ - ]表示+或 - ;
        int[] res = new int[2];//记录系数为x; 系数为常数
        for (String token : tokens){
            if (token.equals("+x") || token.equals("x")){
                res[0] ++;// x表示1x
            }else if (token.equals("-x")){
                res[0] --;
            }else if (token.contains("x")){
                res[0] += Integer.parseInt(token.substring(0,token.length() - 1));
            }else{
                res[1] += Integer.parseInt(token);
            }

        }
        return res;
    }
}

② 在discuss中看到的，思路相同。

class Solution { //9ms
    public String solveEquation(String equation) {
        if(equation == null || equation.length() == 0) return "No solution";
        String[] sides = equation.split("=");
        int[] left = parse(sides[0]), right = parse(sides[1]);
        if(left[0] == right[0]) {
            if(left[1] == right[1]) return "Infinite solutions";
            return "No solution";
        }
        int val = (right[1] - left[1]) / (left[0] - right[0]);
        return "x=" + val;
    }
    
    private int[] parse(String s) {
        int sign = 1, val = 0;
        int[] res = new int[2];
        for(int i = 0; i < s.length(); i ++) {
            char c = s.charAt(i);
            if(c == 'x') {
                if(i == 0) res[0] ++;
                else if(s.charAt(i-1) == '-' || s.charAt(i-1) == '+') res[0] += sign;
                else res[0] += sign * val;
                val = 0;
            } else if(c == '-' || c == '+') {
                res[1] += sign * val;
                val = 0;
                sign = c == '-' ? -1 : 1;
            } else {
                val = val * 10 + c - '0';
            }
        }
        res[1] += sign * val;
        return res;
    }
}