二分法的时间复杂度,直接遍历时间复杂度是。倘若数组的长度是，也就是 4294967296，那么在的最坏的情况下，你要查找4294967296这么多次，如果是用二分的话，最坏查找32次就可以了。
解题模板：
模板只是起到辅助作用，至于真正用的话，还是需要思考如何灵活使用它。
对于有序的数组，可以考虑使用二分，直接暴力遍历，感觉过不了呢。
int start = 0, end = nums.size() - 1; while(start + 1 < end) { int mid = start + (end - start) / 2; if( ... ) { start = mid; } else { end = mid; } //灵活变动，但需要单独判断，可能并行判断，可能有优先级去判断 if（...start....） if（...end....） } 0.第一个错误版本(leetcode 278)
// Forward declaration of isBadVersion API. bool isBadVersion(int version); class Solution { public: int firstBadVersion(int n) { int start = 1, end = n; while(start+1 < end) { int mid = start + (end - start) / 2; if(!isBadVersion(mid)) { start = mid; } else { end = mid; } } if(isBadVersion(start)) { return start; } return end; } }; 1.在排序数组中查找元素的第一个和最后一个位置（leetcode 34）
class Solution { public: vector<int> searchRange(vector<int>& nums, int target) { //for(int i = 0; i < 6; i++)printf("%d ",nums[i]);printf("\n"); vector<int> ret(2,-1); int size = nums.size(); if(size == 0) return ret; // find front int start = 0, end = size - 1; while(start+1 < end) { int mid = start + (end - start) / 2; //printf("start is %d, end is %d, mid is %d \n",start,end,mid); if(nums[mid] >= target) { end = mid; } else { start = mid; } } if(nums[start] == target) { ret[0] = start; } else if(nums[end] == target) { ret[0] = end; } //find back start = 0, end = size - 1; while(start+1 < end) { int mid = start + (end - start) / 2; //printf("start is %d, end is %d, mid is %d \n",start,end,mid); if(nums[mid] <= target) { start = mid; } else { end = mid; } } if(nums[end] == target) { ret[1] = end; } else if(nums[start] == target) { ret[1] = start; } return ret; } }; 2.搜索二维矩阵（leetcod 74）
class Solution { public: bool searchMatrix(vector<vector<int>>& matrix, int target) { if(matrix.empty() || matrix[0].empty()) return false; int row = matrix.size(); int startRow = 0, endRow = row - 1; while(startRow + 1 < endRow) { int mid = startRow + (endRow - startRow)/2; if(matrix[mid][0] == target) { return true; } else if(matrix[mid][0] < target) { startRow = mid; } else { endRow = mid; } } int selectRow = 0; if(matrix[endRow][0] <= target) { selectRow = endRow; } else { selectRow = startRow; } int startCol = 0, endCol = matrix[0].size() - 1; while(startCol+1 < endCol) { int mid = startCol + (endCol - startCol)/2; if(matrix[selectRow][mid] == target) { return true; } else if(matrix[selectRow][mid] < target) { startCol = mid; } else { endCol = mid; } } if(matrix[selectRow][startCol] == target || matrix[selectRow][endCol] == target) return true; return false; } }; 3.搜索二维矩阵 III（leetcod 240）
class Solution { public: bool searchMatrix(vector<vector<int> >& matrix, int target) { if(matrix.empty() || matrix[0].empty()) return false; int numOfFind = matrix.size() < matrix[0].size() ? matrix.size():matrix[0].size(); //printf("numOfFind is %d\n",numOfFind); for(int i = 0; i < numOfFind; i++) { //printf("i is %d\n",i); bool upDown = find(matrix, i, target, true); bool leftRight = find(matrix, i, target, false); //printf("upDown is %d, leftRight is %d\n",upDown,leftRight); if(upDown || leftRight) return true; } return false; } private: bool find(vector<vector<int> >& matrix, int i, int target, bool isFindRow) { int start = i, end = isFindRow?matrix.size()-1:matrix[0].size()-1; //printf("isFindRow is %d\n",isFindRow); while(start + 1 < end) { int mid = start + (end - start)/2; //printf("start is %d, end is %d, mid is %d\n", start, end, mid); if(isFindRow) { if(matrix[mid][i] < target) { start = mid; } else if(matrix[mid][i] > target) { end = mid; } else { return true; } } else { if(matrix[i][mid] < target