lintcode13- strStr- easy

For a given source string and a target string, you should output the first index(from 0) of target string in source string.

If target does not exist in source, just return -1.

Example

If source = "source" and target = "target", return -1.

If source = "abcdabcdefg" and target = "bcd", return 1.

二重循环，判断一下交错时的下标特点，提早排除。

public class Solution {
    /*
     * @param source: source string to be scanned.
     * @param target: target string containing the sequence of characters to match
     * @return: a index to the first occurrence of target in source, or -1  if target is not part of source.
     */
    public int strStr(String source, String target) {
        // write your code here
        if (source == null || target == null){
            return -1;
        }

        for (int sPnt = 0; sPnt < source.length() - target.length() + 1; ++sPnt){
            int tPnt = 0;
            for (tPnt = 0; tPnt < target.length(); ++tPnt){
                if (sPnt + tPnt >= source.length()){
                    return -1;
                }
                else if (source.charAt(sPnt + tPnt) != target.charAt(tPnt)){
                    break;
                }
            }
            if (tPnt == target.length()){
                return sPnt;
            }
        }

        return -1;

    }
}

1. 避免不必要的对比：写for (int sPnt = 0; sPnt < source.length() - target.length() + 1; ++sPnt){

2.如果你for循环的循环变量之后还要用，就在外面定义它，然后在for循环里给它赋值！！这样for结束后还能用。

int tPnt = 0;
for (tPnt = 0; tPnt < target.length(); ++tPnt){

3.考虑输入““””问题，字符串问题里“” 和null还是不一样的。

4.“”的length()为0

5.“”不能做charAt(0)操作。