二分模板
1 int searchRange(vector<int>& nums, int target) { 2 int l=0,r=nums.size()-1; 3 while(l<r){ 4 int mid=l+r>>1; 5 if(check(mid))r=mid; 6 else l=mid+1; 7 } 8 return l; 9 }
来个题：
看见一个ac代码，可以用1~126的二分去找那个差值（枚举差值呗）
1 class Solution { 2 int m, n; 3 vector<vector<int>> dir = {{1,0},{0,1},{0,-1},{-1,0}}; 4 public: 5 int minimumEffortPath(vector<vector<int>>& heights) { 6 m = heights.size(), n = heights[0].size(); 7 int l = 0, r = 1e6, mid, ans = 0; 8 while(l <= r) 9 { 10 mid = l + ((r-l)>>1); 11 vector<vector<bool>> vis(m, vector<bool>(n,false)); 12 if(dfs(heights, vis, 0, 0, mid)) 13 { 14 r = mid-1; 15 ans = mid; 16 } 17 else 18 l = mid+1; 19 } 20 return ans; 21 } 22 bool dfs(vector<vector<int>>& heights, vector<vector<bool>>& vis, int x, int y, int d) 23 { 24 if(x == m-1 && y == n-1) 25 return true; 26 vis[x][y] = true; 27 int i, j, k; 28 for(k = 0; k < 4; k++) 29 { 30 i = x + dir[k][0]; 31 j = y + dir[k][1]; 32 if(i>=0 && i<m && j>=0 && j<n && !vis[i][j] 33 && abs(heights[x][y]-heights[i][j]) <= d) 34 { 35 if(dfs(heights, vis, i, j, d)) 36 return true; 37 } 38 } 39 return false; 40 } 41 };