1.开两个栈,矩阵栈用于保存矩阵信息,操作符栈用于保存括号
2.遍历计算表达式,遇到A-Z压入矩阵栈,遇到 ( 压入操作符栈,遇到 ) 进行计算
3.计算时从矩阵栈中弹出两个矩阵,计算并记录计算次数,再将计算后的结果矩阵压入矩阵栈中
- public int calculateMatrixCount(List
list, String express) { - Map
indexToMatrixMap = new HashMap<>(); - for (int i = 0; i < list.size(); i++) {
- // 构建矩阵索引
- indexToMatrixMap.put(i, list.get(i));
- }
-
- Stack
charStack = new Stack<>(); - Stack
matrixStack = new Stack<>(); -
- int sum = 0;
- for (char c : express.toCharArray()) {
- if (c >= 'A' && c <= 'Z') {
- matrixStack.push(c+"");
- } else {
- if (c == '(') {
- charStack.push(c+"");
- } else {
- // 栈的顺序是相反的,所以先出栈的是矩阵2
- String matrix2 = getMatrix(matrixStack.pop(), indexToMatrixMap);
- String matrix1 = getMatrix(matrixStack.pop(), indexToMatrixMap);
-
- String[] arr1 = matrix1.split(" ");
- String[] arr2 = matrix2.split(" ");
-
- sum += Integer.parseInt(arr1[0]) * Integer.parseInt(arr1[1]) * Integer.parseInt(arr2[1]);
- String newMatrix = arr1[0] + " " + arr2[1];
- // 将计算结果矩阵压入栈中
- matrixStack.push(newMatrix);
- // 弹出外侧的 (
- charStack.pop();
- }
- }
- }
-
- return sum;
- }
-
- private String getMatrix(String s, Map
indexToMatrixMap) { - if (s.length() > 1) {
- return s;
- }
-
- int index = s.toCharArray()[0] - 'A';
- return indexToMatrixMap.get(index);
- }
时间复杂度:O(n)