LeetCode 2109. Adding Spaces to a String

原题链接在这里：https://leetcode.com/problems/adding-spaces-to-a-string/description/

题目：

You are given a 0-indexed string s and a 0-indexed integer array spaces that describes the indices in the original string where spaces will be added. Each space should be inserted before the character at the given index.

• For example, given s = "EnjoyYourCoffee" and spaces = [5, 9], we place spaces before 'Y' and 'C', which are at indices 5 and 9 respectively. Thus, we obtain "Enjoy Your Coffee".

Return the modified string after the spaces have been added. 

Example 1:

Input: s = "LeetcodeHelpsMeLearn", spaces = [8,13,15]
Output: "Leetcode Helps Me Learn"
Explanation: 
The indices 8, 13, and 15 correspond to the underlined characters in "LeetcodeHelpsMeLearn".
We then place spaces before those characters.

Example 2:

Input: s = "icodeinpython", spaces = [1,5,7,9]
Output: "i code in py thon"
Explanation:
The indices 1, 5, 7, and 9 correspond to the underlined characters in "icodeinpython".
We then place spaces before those characters.

Example 3:

Input: s = "spacing", spaces = [0,1,2,3,4,5,6]
Output: " s p a c i n g"
Explanation:
We are also able to place spaces before the first character of the string.

Constraints:

• 1 <= s.length <= 3 * 105
• s consists only of lowercase and uppercase English letters.
• 1 <= spaces.length <= 3 * 105
• 0 <= spaces[i] <= s.length - 1
• All the values of spaces are strictly increasing.

题解：

Traversae the string from left to right, when index == spaces[ind], then add a space to res StringBuilder and ind++. 

Note: when checking spaces[ind], check ind < spaces.length first.

Time Complexity: O(n + m). n = s.length(). m = spaces.length.

Space: O(n + m)

AC Java:

class Solution {
    public String addSpaces(String s, int[] spaces) {
        if(s == null || spaces == null || spaces.length == 0){
            return s;
        }

        StringBuilder sb = new StringBuilder();
        int ind = 0;
        for(int i = 0; i < s.length(); i++){
            if(ind < spaces.length && i == spaces[ind]){
                sb.append(" ");
                ind++;
            }

            sb.append(s.charAt(i));
        }

        return sb.toString();
    }
}