String源码理解之indexOf、lastIndexOf函数

1前言

不多说，直接上源码

2indexOf源码

我自己的理解，可能表述不清，多看几遍，不行就debug跟一遍代码自然就懂了。
/**
* Code shared by String and StringBuffer to do searches. The
* source is the character array being searched, and the target
* is the string being searched for.
*
* @param source the characters being searched. 要搜索的源字符串
* @param sourceOffset offset of the source string. 源字符串偏移量即起始位置
* @param sourceCount count of the source string. 源字符串长度
* @param target the characters being searched for. 要搜索的目标字符串
* @param targetOffset offset of the target string. 目标字符串偏移量即起始位置
* @param targetCount count of the target string. 目标字符串长度
* @param fromIndex the index to begin searching from. 开始搜索位置
*/
static int indexOf(char[] source, int sourceOffset, int sourceCount,
char[] target, int targetOffset, int targetCount,
int fromIndex) {

        if (fromIndex >= sourceCount) {
            return (targetCount == 0 ? sourceCount : -1);
        }
        if (fromIndex < 0) {
            fromIndex = 0;
        }
        if (targetCount == 0) {
            return fromIndex;
        }

        //开始位置
        char first = target[targetOffset];
        //要循环的最大次数
        int max = sourceOffset + (sourceCount - targetCount);

        for (int i = sourceOffset + fromIndex; i <= max; i++) {
            /* Look for first character. */
        	//找到第一个字符的位置，for循环嵌套while循环直到找到
            if (source[i] != first) {
                while (++i <= max && source[i] != first);
            }

            /* Found first character, now look at the rest of v2 */
            if (i <= max) {
            	//第二个字符
                int j = i + 1;
                int end = j + targetCount - 1;
                //第一个字符之后的字符是否也一一相等
                for (int k = targetOffset + 1; j < end && source[j]
                        == target[k]; j++, k++);

                //如果j=end则说明第一个字符之后的字符也一一相等
                
                //否则继续循环找下一个字符
                if (j == end) {
                    /* Found whole string. */
                    return i - sourceOffset;
                }
            }
        }
        return -1;
    }

3lastindexOf源码

/**
* Code shared by String and StringBuffer to do searches. The
* source is the character array being searched, and the target
* is the string being searched for.
*
* @param source the characters being searched.
* @param sourceOffset offset of the source string.
* @param sourceCount count of the source string.
* @param target the characters being searched for.
* @param targetOffset offset of the target string.
* @param targetCount count of the target string.
* @param fromIndex the index to begin searching from.
/
static int lastIndexOf(char[] source, int sourceOffset, int sourceCount,
char[] target, int targetOffset, int targetCount,
int fromIndex) {
/
* Check arguments; return immediately where possible. For
* consistency, don't check for null str.
/
int rightIndex = sourceCount - targetCount;
if (fromIndex < 0) {
return -1;
}
if (fromIndex > rightIndex) {
fromIndex = rightIndex;
}
/ Empty string always matches. */
if (targetCount == 0) {
return fromIndex;
}

    int strLastIndex = targetOffset + targetCount - 1;
    char strLastChar = target[strLastIndex];
    int min = sourceOffset + targetCount - 1;
    int i = min + fromIndex;

    //一个标记循环，普通的continue直接跳出当前循环
    startSearchForLastChar:
    while (true) {
    	//哈哈哈  为啥这个地方的源码没有注释呢，程序员肯定是第一遍思考indeof的时候也头疼，所以注释辅助思考，遇到lastindexof他自然就不用思考太多，因为与indexof大同小异。
    	//从后往前找，找到最后一个相等的字符
        while (i >= min && source[i] != strLastChar) {
            i--;
        }
        if (i < min) {
            return -1;
        }
        int j = i - 1;
        int start = j - (targetCount - 1);
        int k = strLastIndex - 1;

        //循环直到j=start，说明从最后一个字符往前的所有字符都匹配
        while (j > start) {
            if (source[j--] != target[k--]) {
                i--;
                continue startSearchForLastChar;
            }
        }
        return start - sourceOffset + 1;
    }
}

4总结

亮点：
1、匹配首个目标字符。for循环中，嵌套while循环，从源字符串中，找到匹配目标字符的位置。
2、小范围匹配目标字符串。for循环中，嵌套for循环，在[目标字符串]长度范围内，遍历匹配，如果全部匹配，则return。