两道关于算法的面试题

1, 给定一个字符串, 已知字符串只包含'(', ')', '[', ']'四种括号字符, 要求: 写一个程序, 输入随意的字符串, 判断括号是否匹配

例: 

输入: "()"

返回: "true"

输入: "(]"

返回: "false"

输入: "()[()]"

返回: "true"

输入: "()[()()]([][])[()[[[[]]]]]"

返回: "true"

package test1;

import java.util.ArrayList;
import java.util.List;

public class TestStack {

    private List<Character> elements = new ArrayList<>();

    private void in(Character o) {
        this.elements.add(o);
    }

    private Character out() {
        int length = this.elements.size();
        if (length > 0) {
            return elements.remove(length - 1);
        }
        return null;
    }

    public int size() {
        return this.elements.size();
    }

    public boolean valid(String str) {
        this.elements.clear();
        if (str != null && str.length() > 0) {
            char[] chars = str.toCharArray();
            for (char c : chars) {
                char left1 = '(';
                if (c == left1) {
                    this.in(c);
                }
                char right1 = '[';
                if (c == right1) {
                    this.in(c);
                }
                char left2 = ')';
                if (c == left2) {
                    char outElement = this.out();
                    boolean r = outElement == left1;
                    if (!r) {
                        return false;
                    }
                }
                char right2 = ']';
                if (c == right2) {
                    char outElement = this.out();
                    boolean r = outElement == right1;
                    if (!r) {
                        return false;
                    }
                }
            }
        }
        return this.elements.size() == 0;
    }
}

package test1;

public class TestDemo {
    public static void main(String[] args) {
        String str1 = "()";
        String str2 = "[]";
        String str3 = "(]";
        String str4 = "((()";
        String str5 = "()[()]";
        String str6 = "([)]";
        String str7 = "()[()()]([][])[()[[[[]]]]]";

        TestStack testStack = new TestStack();
        System.out.println(str1 + ": " + testStack.valid(str1));
        System.out.println(str2 + ": " + testStack.valid(str2));
        System.out.println(str3 + ": " + testStack.valid(str3));
        System.out.println(str4 + ": " + testStack.valid(str4));
        System.out.println(str5 + ": " + testStack.valid(str5));
        System.out.println(str6 + ": " + testStack.valid(str6));
        System.out.println(str7 + ": " + testStack.valid(str7));
    }
}

 判断括号匹配的另一种解法

public class Main {

    private static String f1 = "(}";
    private static String f2 = "{)";

    public static void main(String[] args) {
        String src = "({((){()()}((){}){()()})})";
        System.out.println(valid(src));
    }

    public static boolean valid(String src) {
        if ("".equals(src)) {
            return false;
        }
        return _valid(src);
    }

    public static boolean _valid(String src) {
        if (src != null && src.trim().length() > 0) {
            if (src.trim().length() % 2 != 0) {
                return false;
            }
            if (!src.contains(f1) && !src.contains(f2)) {
                src = src.replace("()", "").replace("{}", "");
                return valid(src);
            }
            return false;
        }
        return true;
    }
}

　　

2, 给定两个数组, 两个数组长度大于0(两个数组的长度可能相同也可能不同), 并且各自已经按照由小到大排序, 要求: 合并两个数组形成一个新的数组, 并且不需要重新排序, 新的数组也按照由小到大排列

package test2;

import java.util.Arrays;

public class TestDemo {
    public static void main(String[] args) {
        int[] arr1 = {1, 3, 5, 9, 15};
        int[] arr2 = {2, 5, 7, 8, 12, 13, 19};
//        int[] arr1 = {1, 2, 3, 4};
//        int[] arr2 = {1, 2, 3};
//        int[] arr1 = {-1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1};
//        int[] arr2 = {1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1};

        int[] newArr = new int[arr1.length + arr2.length];
        int point1 = 0;
        int point2 = 0;

        for (int i = 0; i < newArr.length; i++) {
            if (point1 >= arr1.length && point2 < arr2.length) {
                newArr[i] = arr2[point2++];
                continue;
            }
            if (point1 < arr1.length && point2 >= arr2.length) {
                newArr[i] = arr1[point1++];
                continue;
            }
            int result = Integer.compare(arr1[point1], arr2[point2]);
            switch (result) {
                case -1:
                    newArr[i] = arr1[point1];
                    point1++;
                    break;
                case 0:
                    newArr[i++] = arr1[point1];
                    newArr[i] = arr2[point2];
                    point1++;
                    point2++;
                    break;
                case 1:
                    newArr[i] = arr2[point2];
                    point2++;
                    break;
            }
        }
        System.out.println(Arrays.toString(newArr));
    }
}