Implement strStr() & Largest Rectangle in Histogram & Maximal Rectangle

(1) Implement strStr() 

暴力法:

class Solution {
public:
    char *strStr(char *haystack, char *needle) {
        if(!haystack || !needle)
            return NULL;
        int m=strlen(haystack);
        int n=strlen(needle);
        int j;
        
        for(int i=0;i<=m-n;i++)
        {
            for(j=0;j<n;j++)
            {
                if(haystack[i+j]!=needle[j])
                    break;
            }
            if(j==n)
                return haystack+i;
        }
        return NULL;
    }
};

还有其他方法Boyer Moore法、KMP法[1]。

(2) Largest Rectangle in Histogram

原理见[2]，非常详细：

class Solution {
public:
    int largestRectangleArea(vector<int> &height) {
        if(height.size()==0)
            return 0;
        stack<int> stk;
        
        int i=0,maxarea=0;
        height.push_back(0);
        while(i<height.size())
        {
            if(stk.empty() || height[stk.top()]<=height[i])
                stk.push(i++);
            else
            {
               int t=stk.top();
               stk.pop();
               int tmp=height[t]*(stk.empty() ? i : i-stk.top()-1);
               maxarea=max(maxarea,tmp);
            }
        }
        
        return maxarea;
    }
};

(3) Maximal Rectangle

在(2)的基础上[3]：

class Solution {
private:
    int largestRectangleArea(vector<int> &height) {
        if(height.size()==0)
            return 0;
        stack<int> stk;
        
        int i=0,maxarea=0;
        height.push_back(0);
        while(i<height.size())
        {
            if(stk.empty() || height[stk.top()]<=height[i])
                stk.push(i++);
            else
            {
               int t=stk.top();
               stk.pop();
               int tmp=height[t]*(stk.empty() ? i : i-stk.top()-1);
               maxarea=max(maxarea,tmp);
            }
        }
        
        return maxarea;
    }

public:
    int maximalRectangle(vector<vector<char> > &matrix) {
        int m=matrix.size();
        if(m==0)
            return 0;
        int n=matrix[0].size();
        if(n==0)
            return 0;
        int maxarea=0;
        
        vector<vector<int>> height;
        height.resize(m);
        for(int i=0;i<m;i++)
            height[i].resize(n);
        
        for(int i=0;i<m;i++)
            for(int j=0;j<n;j++)
            {
                if(matrix[i][j]=='0')
                    height[i][j]=0;
                else
                    height[i][j]=(i==0 ? 1 : height[i-1][j]+1);
            }
        
        for(int i=0;i<m;i++)
            maxarea=max(maxarea,largestRectangleArea(height[i]));
            
        return maxarea;
    }
};

参考：

[1] http://blog.youkuaiyun.com/kenden23/article/details/17029625

[2] http://www.cnblogs.com/lichen782/p/leetcode_Largest_Rectangle_in_Histogram.html

[3] http://www.cnblogs.com/lichen782/p/leetcode_maximal_rectangle.html