翻轉字符串的相關題目

LeetCode: 557. Reverse Words in a String III

一、LeetCode: 557. Reverse Words in a String III

題目描述

Given a string, you need to reverse the order of characters in each word within a sentence while still preserving whitespace and initial word order.

Example 1:

Input: “Let’s take LeetCode contest”

Output: “s’teL ekat edoCteeL tsetnoc”

Note: In the string, each word is separated by single space and there will not be any extra space in the string.

空格分隔，逐個反轉

class Solution {
    public String reverseWords(String s) {

        String[] ss = s.split(" ");
        StringBuilder sb = new StringBuilder();
        int n = ss.length;

        for (int i = 0; i < n - 1; i++) {
            sb.append(reverse(ss[i]) + " ");
        }
        sb.append(reverse(ss[n - 1]));

        return sb.toString();
    }

    public String reverse(String str) {
        StringBuilder sb = new StringBuilder(str);
        return sb.reverse().toString();
    }
}

LeetCode: 541. Reverse String II

二、LeetCode: 541. Reverse String II

題目描述

Given a string and an integer k, you need to reverse the first k characters for every 2k characters counting from the start of the string. If there are less than k characters left, reverse all of them. If there are less than 2k but greater than or equal to k characters, then reverse the first k characters and left the other as original.

Example:

Input: s = “abcdefg”, k = 2

Output: “bacdfeg”

Restrictions:

The string consists of lower English letters only.

Length of the given string and k will in the range [1, 10000]

class Solution {
    public String reverseStr(String s, int k) {

        int n = s.length();
        int i = 0, j = n - 1;
        char[] c = s.toCharArray();

        while (i < n) {
            j = Math.min(i + k - 1, n - 1);
            reverse(c, i, j);
            i += 2 * k;
        }

        return new String(c);

    }

    public void reverse(char[] c, int i, int j) {
        while (i < j) {
            char temp = c[i];
            c[i] = c[j];
            c[j] = temp;
            i++;
            j--;
        }
    }
}

LeetCode: 344. Reverse String

三、LeetCode: 344. Reverse String

題目描述

Write a function that takes a string as input and returns the string reversed.

Example:

Given s = “hello”, return “olleh”.

當然也可以用 StringBuilder 的 reverse 做，不過用雙指針更快。

class Solution {
    public String reverseString(String s) {

        int len = s.length();
        int left = 0, right = len - 1;
        char[] c = s.toCharArray();

        while(left < right) {
            char temp = c[left];
            c[left] = c[right];
            c[right] = temp;
            left++;
            right--;
        }

        return new String(c);

    }
}

LeetCode: 345. Reverse Vowels of a String

四、LeetCode: 345. Reverse Vowels of a String

題目描述

Write a function that takes a string as input and reverse only the vowels of a string.

Example 1:

Given s = “hello”, return “holle”.

Example 2:

Given s = “leetcode”, return “leotcede”.

Note:

The vowels does not include the letter “y”.

同樣地，雙指針，注意考慮 vowel 的大小寫

class Solution {
    public String reverseVowels(String s) {
        // a e i o u
        int len = s.length();
        int left = 0, right = len - 1;
        char[] c = s.toCharArray();

        while(left < right) {
            if (isVowel(c[left]) && isVowel(c[right])) {
                char temp = c[left];
                c[left] = c[right];
                c[right] = temp;
                left++;
                right--;
            } else if (!isVowel(c[left]))
                left++;
            else right--;
        }

        return new String(c);
    }

    public boolean isVowel(char c) {
        return c == 'a' || c == 'A'
           || c == 'e' || c == 'E'
           || c == 'i' || c == 'I'
           || c == 'o' || c == 'O'
           || c == 'u' || c == 'U';
    }
}

LeetCode: 11. Container With Most Water

五、LeetCode: 11. Container With Most Water

題目描述

Given n non-negative integers a1, a2, …, an, where each represents a point at coordinate (i, ai). n vertical lines are drawn such that the two endpoints of line i is at (i, ai) and (i, 0). Find two lines, which together with x-axis forms a container, such that the container contains the most water.

Note: You may not slant the container and n is at least 2.

首先理解題意，就是找一個最大的容器容納最多的水，這個容器由兩個板子組成，每個 an 在數組中的值就是這個板子的高度，畫一下圖就更清晰了。

[2,1,2] 模擬畫圖 ≡ω≡

|   |
| | |     板子...
--------- x軸
0 1 2     數組下標

然後抽象出來就是求 (j - i) * Math.min(height[i], height[j]) 的最大值

暴力算法很容易想到，但是會超時

有一種 O(n) 的做法，但是不是很好理解，discuss 裏有比較清楚的講解 Yet another way to see what happens in the O(n) algorithm

我理解的核心思想是突破那個短板，如果左邊的短，就移動左邊的板子，如果右邊的短，就移動右邊的板子，看能不能形成更大的容器。

class Solution {
    public int maxArea(int[] height) {

        int n = height.length;
        int max = 0;
        int i = 0, j = n - 1;

        while (i < j) {
            max = Math.max(max, Math.min(height[i], height[j]) * (j - i));
            if (height[i] < height[j])
                i++;
            else
                j--;
        }

        return max;
    }
}

翻轉字符串的相關題目