二分查找法（递归与循环实现）

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问题：

给定一个排序数组和一个数k，要求找到第一个k的位置和最后一个k的位置

解析：

由于给定的数组是从小到大排序的，故可以按照二分查找法来找，下面分别从递归和循环两种方法来阐述：

//递归方法int GetFirstK(int* data, int length, int k, int start, int end){if (start > end)return -1;int middleindex = (start + end) / 2;int middledata = data[middleindex];if (middledata>k){end = middleindex - 1;}else if (middledata<k){start = middleindex + 1;}else{if (middleindex==0||(middleindex>0&&data[middleindex-1]!=k))//判断左边的元素是否等于k，若等于，说明第一个k在左边,否则第一个k就是middledata{return middleindex;}else{end = middleindex - 1;}}return  GetFirstK(data, length, k,start, end);}

//循环法int GetFirstK(int* data, int length, int k){int start = 0;int end = length - 1;while (start<=end){int middleindex = (start + end) / 2;int middledata = data[middleindex];if (middledata<k){start = middleindex + 1;}else if (middleindex>k){end = middleindex - 1;}else{if (middleindex==0||(middleindex>0&&data[middleindex-1]!=k)){return middleindex;}else{end = middleindex - 1;}}}return -1;}

//递归法int GetLastK(int* data, int length, int k, int start, int end){if (start>end){return -1;}int middleindex = (start + end) >> 1;int middledata = data[middleindex];if (middledata<k){start = middleindex + 1;}else if (middledata>k){end = middleindex - 1;}else{if (middleindex==length-1||(middleindex<length-1&&data[middleindex+1]!=k)){return middleindex;}else{start = middleindex + 1;}}return GetLastK(data, length, k, start, end);}

//循环法int GetLastK(int* data, int length, int k){int start = 0;int end = length - 1;while (start<=end){int middleindex = (start + end) >> 1;int middledata = data[middleindex];if (middledata>k){end = middleindex - 1;}else if (middledata<k){start = middleindex + 1;}else{if (middleindex==length-1||(middleindex<length-1&&data[middleindex+1]!=k)){return middleindex;}else{start = middleindex + 1;}}}return -1;}

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