Implement strStr()

LeetCode

注意substring()的用法：

public String substring(int startIndex, int endIndex):
This method returns new String object containing the substring of the given string from specified startIndex to endIndex.

startIndex: inclusive
endIndex: exclusive

eg:
String s="hello";
System.out.println(s.substring(0,2));//he
In the above substring, 0 points to h but 2 points to e (because end index is exclusive).

我的解法：

class Solution {
    /**
     * Returns a index to the first occurrence of target in source,
     * or -1  if target is not part of source.
     * @param source string to be scanned.
     * @param target string containing the sequence of characters to match.
     */
    public int strStr(String source, String target) {
        // write your code here
        if (source == null || target == null){
            return -1;
        }
        int lenOfSource = source.length();
        int lenOfTarget = target.length();
        if (lenOfSource < lenOfTarget){
            return -1;
        } else if (lenOfSource == lenOfTarget){
            if (target.equals(source)){
                return 0;
            } else {
                return -1;
            }
        } else if (lenOfSource > lenOfTarget){
            for (int i = 0; i < lenOfSource - lenOfTarget + 1; i++){
                if (target.equals(source.substring(i, i + lenOfTarget))){
                    return i;
                }
            }
        }
        return -1;
    }
}

感觉我的思路很Straightforwawrd, 分为三种情况：
- source的长度比target还小，直接return -1;
- source的长度与target相等，如果包含子串，则与子串相等，返回0；
- source的长度比target大，则依次检查原串是否包含子串；

第三种情况里，注意：

 for (int i = 0; i < lenOfSource - lenOfTarget + 1; i++){
                if (target.equals(source.substring(i, i + lenOfTarget))){
                    return i;
                }
            }

从原字符串的index = 0开始，依次循环到可能包含子串的Index为止。如果只剩下不到子串长度的字符串，则不需要再继续检查了。