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| Line 1: |
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| − | You just need to maintain a count, like so:
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| | | | |
| − | <pre>
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| − | public static boolean isBalanced(String str) {
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| − | int count = 0;
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| | | | |
| − | for (int i = 0, n = str.length(); i < n; i++) {
| + | Back to [[Chapter 4]] |
| − | switch (str.charAt(i)) {
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| − | case '(':
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| − | count++;
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| − | break;
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| − | case ')':
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| − | count--;
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| − | break;
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| − | default:
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| − | throw new IllegalArgumentException();
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| − | }
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| − | | |
| − | if (count < 0) {
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| − | System.out.println("Imbalance at index " + i);
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| − | return false;
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| − | }
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| − | }
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| − | | |
| − | if (count != 0) {
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| − | System.out.println("Imbalance at index " + (str.length() - 1));
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| − | return false;
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| − | }
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| − | | |
| − | return true;
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| − | }
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| − | </pre>
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| − | | |
| − | Another Approach to use stack in-order to solve this problem;
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| − | | |
| − | A. Read the string from left to right as CHR
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| − | B. Keep pushing CHR until CHR is not ")"
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| − | 1) Pop from stack once CHR == ")"
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| − | if poped element is NOT "("
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| − | return FAlse
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| − | Repeat A
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| − | --[[User:Max|Max]] 07:18, 16 June 2010 (EDT)
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| − |
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| − | ----
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| − | The approach below maintains a stack where the index of each opening '(' is pushed onto the stack. As each closing ')' is found, the index of the matching '(' is popped off the stack.
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| − | | |
| − | The string is unbalanced if there:
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| − | * is an attempt to pop an element off an empty stack. The position of the current character can be reported as the position of the first offending parenthesis.
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| − | * are any elements left on the stack. The value of the first element in the stack can be reported as the position of the first offending parenthesis.
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| − | <pre>
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| − | #include <iostream>
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| − | #include <stack>
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| − | #include <string>
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| − | | |
| − | int main (int argc, char* argv[]) {
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| − | if (argc != 2) {
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| − | std::cout << "pass a string to check if it's parens are balanced" << std::endl;
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| − | exit(1);
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| − | }
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| − | | |
| − | std::stack<int> parens;
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| − | std::string input = argv[1];
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| − | | |
| − | for (int i = 0; i < input.length(); i++) {
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| − | char ch = input[i];
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| − | if (ch == ')') {
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| − | if (parens.size() == 0) {
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| − | std::cout << "unmatched ')' at position: " << i << std::endl;
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| − | exit(1);
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| − | }
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| − | parens.pop();
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| − | }
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| − | else if (ch == '(') {
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| − | parens.push(i);
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| − | }
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| − | }
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| − | if (parens.size() != 0) {
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| − | while (parens.size() > 1) {
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| − | parens.pop();
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| − | }
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| − | std::cout << "unmatched '(' at position: " << parens.top() << std::endl;
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| − | exit(1);
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| − | }
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| − | else {
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| − | std::cout << "parens balanced!" << std::endl;
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| − | }
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| − | }
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| − | </pre>
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| − | --[[User:Neonstalwart|Neonstalwart]] ([[User talk:Neonstalwart|talk]]) 22:47, 26 March 2015 (EDT)
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| − | | |
| − | ----
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| − | | |
| − | Max's first example above appears to incorrectly report the last character in the string as the offending index if there is a surplus of opening parentheses. The following solution (in JavaScript) takes a similar approach, but correctly reports the index of the first unmatched open parenthesis (more like Neonstalwart's solution, but without requiring a stack).
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| − | | |
| − | This function returns a number: -1 if balanced, otherwise the index of the first offending parenthesis.
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| − | | |
| − | <pre>
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| − | function checkParens(str) {
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| − | var numOpen = 0;
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| − | var length = str.length;
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| − | var offender;
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| − | | |
| − | for (var i = 0; i < length; i++) {
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| − | if (str[i] === '(') {
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| − | if (!numOpen) {
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| − | // Store index of first open paren in case it is never closed
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| − | offender = i;
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| − | }
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| − | numOpen++;
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| − | }
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| − | else if (str[i] === ')') {
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| − | if (!numOpen) {
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| − | // Closing paren with none open = instant fail
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| − | return i;
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| − | }
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| − | numOpen--;
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| − | }
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| − | }
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| − | | |
| − | return numOpen === 0 ? -1 : offender;
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| − | }
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| − | </pre>
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| − | --[[User:CnEY|CnEY]] ([[User talk:CnEY|talk]]) 23:15, 27 August 2015 (EDT)
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| − | | |
| − | | |
| − | Back to [[Chapter 3]] | |
