标签：style blog color io os ar for sp div
Given a sorted array of integers, find the starting and ending position of a given target value.
Your algorithm‘s runtime complexity must be in the order of O(log n).
If the target is not found in the array, return [-1, -1].
For example,
Given [5, 7, 7, 8, 8, 10] and target value 8,
return [3, 4].
给的那个一个有序数组和一个目标target，找到这个目标在有序数组里的开始和结束的下标。如果不存在就返回 -1 -1。注，如果数组中目标值只有一个，那就返回两个相同的下标。
用binary search找到target 然后再在左边binary search找到左边（包括mid）第一个目标，记录下标。再在右边binary 找到最后一个目标。记录下标。返回。
代码如下：
class Solution { public: vector<int> searchRange(int A[], int n, int target) { vector<int> wr, ans; wr.push_back(-1); // 要学会可以用 vector<int> wr(2,-1);直接将其初始化为想要的 wr.push_back(-1); if(n == 0) return wr; int l = 0, r = n -1, mid = 0; while(l <= r) { mid = (l+r)/2; if (A[mid] < target) // 如果mid小于target那么可以把左边一块去掉 { l = mid + 1; continue; } if (A[mid] > target) // 如果mid大于target那么可以把右边一块去掉 { r = mid -1; continue; } if (A[mid] == target)// mid 找到了刚好，那就肯定在这里面要有return，如果一直没有找到，跳出while后就返回wr { // find start from l to mid，binary search int start = l, l_r = mid; // l_r指mid左边的最右 while(l <= l_r) { int tmp_mid = (l + l_r)/2; if(A[tmp_mid] != target) {l = tmp_mid + 1;} if(A[tmp_mid] == target) {l_r = tmp_mid - 1; start = tmp_mid;} } ans.push_back(start); // find end from mid to r，binary search int r_l = mid, End = mid; // r_l指mid右边的最左 while(r_l <= r) { int tmp_mid = (r_l + r)/2; if(A[tmp_mid] != target) {r = tmp_mid - 1;} if(A[tmp_mid] == target) {r_l = tmp_mid + 1; End = tmp_mid;} } ans.push_back(End); return ans; } } return wr; } };
leetcode第33题--Search for a Range
标签：style blog color io os ar for sp div
原文地址：http://www.cnblogs.com/higerzhang/p/4046972.html
