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找出字符串中第一个匹配项的下标

给你两个字符串 haystack 和 needle ，请你在 haystack 字符串中找出 needle 字符串的第一个匹配项的下标（下标从 0 开始）。如果 needle 不是 haystack 的一部分，则返回 -1 。

示例 1：
输入：haystack = “sadbutsad”, needle = “sad”
输出：0
解释：“sad” 在下标 0 和 6 处匹配。
第一个匹配项的下标是 0 ，所以返回 0 。

方法一、暴力法

类似双指针的思想，遍历解决。优势就是简单容易理解，但时间复杂度不是最优。

复杂度
时间复杂度：O(n*m)
空间复杂度：O(1)

Swift

func strStr(_ haystack: String, _ needle: String) -> Int {let cntL = haystack.countlet cntR = needle.countvar i = 0while i+cntR <= cntL {var flag = truefor j in 0..<cntR {if haystack[haystack.index(haystack.startIndex, offsetBy: i+j)] != needle[needle.index(needle.startIndex, offsetBy: j)] {flag = falsebreak}}if flag {return i}i += 1}return -1}

OC

-(NSInteger)strStr:(NSString *)haystack needle:(NSString *)needle {if (needle.length <= 0) {return -1;}NSInteger n = haystack.length, m = needle.length;for (NSInteger i=0; i+m<=n; i++) {NSInteger j = 0;while (j<m && [haystack characterAtIndex:i+j] == [needle characterAtIndex:j]) {j++;}if (j == m) {return i;}}return -1;
}

方法二、KMP算法（推荐）

面试的时候如果能撸出来，那么基本就稳了，相反，如果只知道暴力法，那可能结果就是回去等通知吧。
具体算法讲解请参考：
彻底搞懂KMP算法！！（配视频讲解）

Swift

//KMP 解法func strStr(_ haystack: String, _ needle: String) -> Int {let n = haystack.countlet m = needle.countif m == 0 {return 0}//构造next数组var pi = Array(repeating: 0, count: m+1)var j = 0for i in 1..<m {while j > 0 && needle[needle.index(needle.startIndex, offsetBy: i)] != needle[needle.index(needle.startIndex, offsetBy:j)] {j = pi[j-1]}if needle[needle.index(needle.startIndex, offsetBy: i)] == needle[needle.index(needle.startIndex, offsetBy: j)]  {j += 1}pi[i] = j}j = 0for i in 0..<n {while j>0 && haystack[haystack.index(haystack.startIndex, offsetBy: i)] != needle[needle.index(needle.startIndex, offsetBy: j)] {j = pi[j-1]}if haystack[haystack.index(haystack.startIndex, offsetBy: i)] == needle[needle.index(needle.startIndex, offsetBy: j)] {j += 1}if j == m {return i-m+1}}return -1}

OC

//KMP
-(NSInteger)strStr:(NSString *)haystack needle:(NSString *)needle {NSInteger n = haystack.length;NSInteger m = needle.length;if (m <= 0) {return 0;}NSMutableArray *next = [NSMutableArray array];for (NSInteger i=0; i<m; i++) {next[i] = @(0);}//求next数组NSInteger i=1;NSInteger j = 0;for (; i<m; i++) {//不相等的情况if (j>0 && [needle characterAtIndex:i] != [needle characterAtIndex:j]) {j = [next[j-1] integerValue];}//相等情况if ([needle characterAtIndex:i] == [needle characterAtIndex:j]) {j++;}//赋值next[i] = [NSNumber numberWithInteger:j];}j = 0;for (i=0; i<n; i++) {if (j > 0 && [haystack characterAtIndex:i] != [needle characterAtIndex:j]) {j = [next[j-1] integerValue];}if ([haystack characterAtIndex:i] == [needle characterAtIndex:j]) {j++;}if (j == m) {return i-m+1;}}return -1;
}