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640. Solving Equations: Simple Simulation Problems
2022-08-10 14:53:00 【Water palace trifoliate brush diary】
题目描述
这是 LeetCode 上的 640. 求解方程 ,难度为 中等.
Tag : 「模拟」、「数学」、「双指针」
求解一个给定的方程,将 x
以字符串 "x=#value"
的形式返回.该方程仅包含 '+'
, '-'
操作,变量 x
和其对应系数.
如果方程没有解,请返回 "No solution"
.如果方程有无限解,则返回 "Infinite solutions"
.
题目保证,如果方程中只有一个解,则 'x' 的值是一个整数.
示例 1:
输入: equation = "x+5-3+x=6+x-2"
输出: "x=2"
示例 2:
输入: equation = "x=x"
输出: "Infinite solutions"
示例 3:
输入: equation = "2x=x"
输出: "x=0"
提示:
equation
只有一个'='
.equation
方程由整数组成,其绝对值在 范围内,不含前导零和变量'x'
.
模拟
为了方便,我们令 equation
为 s
.
Since the operator only +
和 -
,Therefore, there is no need to consider the operation priority,Calculations can be performed during the traversal process.
使用变量 x
和 num
Respectively refer to the current operation result The coefficients and numerical part of ,从前往后处理 s
的每个字符,Discuss on a case-by-case basis according to character type,Suppose the currently processed value is :
若 +/-
:At this time, it affects the positive or negative of the next operand value,修改对应的op
标识;若 数值
:At this time, the complete operation value is fetched(The operand value may be about 的描述,May be purely numeric),Assume continuous segments between is the current operation value,根据 是否为字符x
可知,是要将 The value is accumulated to the variablex
,还是将 The value is accumulated to the variablenum
;若 =
:This represents that the left side of the equation has been processed,将变量x
和num
进行翻转(The meaning is to move the result of the operation on the left to the right),and continue processing.
when the entire string s
处理完后,We get finally about 的系数 x
,and numerical size num
.
根据 x
是否为 know the answer:
若 x
为 :此时根据num
是否为 可知是Infinite solutions
(对应num
为 ) 还是No solution
(对应num
不为 )若 x
不为 :对x
和num
After making an appointment,Return the corresponding answer.
Java 代码:
class Solution {
public String solveEquation(String s) {
int x = 0, num = 0, n = s.length();
char[] cs = s.toCharArray();
for (int i = 0, op = 1; i < n; ) {
if (cs[i] == '+') {
op = 1; i++;
} else if (cs[i] == '-') {
op = -1; i++;
} else if (cs[i] == '=') {
x *= -1; num *= -1; op = 1; i++;
} else {
int j = i;
while (j < n && cs[j] != '+' && cs[j] != '-' && cs[j] != '=') j++;
if (cs[j - 1] == 'x') x += (i < j - 1 ? Integer.parseInt(s.substring(i, j - 1)) : 1) * op;
else num += Integer.parseInt(s.substring(i, j)) * op;
i = j;
}
}
if (x == 0) return num == 0 ? "Infinite solutions" : "No solution";
else return "x=" + (num / -x);
}
}
TypeScript 代码:
function solveEquation(s: string): string {
let x = 0, num = 0, n = s.length
for (let i = 0, op = 1; i < n; ) {
if (s[i] == '+') {
op = 1; i++;
} else if (s[i] == '-') {
op = -1; i++
} else if (s[i] == '=') {
x *= -1; num *= -1; op = 1; i++;
} else {
let j = i
while (j < n && s[j] != '+' && s[j] != '-' && s[j] != '=') j++
if (s[j - 1] == 'x') x += (i < j - 1 ? Number(s.substring(i, j - 1)) : 1) * op
else num += Number(s.substring(i, j)) * op
i = j
}
}
if (x == 0) return num == 0 ? "Infinite solutions" : "No solution"
else return "x=" + (num / -x)
};
时间复杂度: 空间复杂度:使用 charAt
替换toCharArray
.复杂度为
最后
这是我们「刷穿 LeetCode」系列文章的第 No.640
篇,系列开始于 2021/01/01,截止于起始日 LeetCode 上共有 1916 道题目,部分是有锁题,我们将先把所有不带锁的题目刷完.
在这个系列文章里面,除了讲解解题思路以外,还会尽可能给出最为简洁的代码.如果涉及通解还会相应的代码模板.
为了方便各位同学能够电脑上进行调试和提交代码,我建立了相关的仓库:https://github.com/SharingSource/LogicStack-LeetCode .
在仓库地址里,你可以看到系列文章的题解链接、系列文章的相应代码、LeetCode 原题链接和其他优选题解.
更多更全更热门的「笔试/面试」相关资料可访问排版精美的 合集新基地
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