Naive solution is O(n^4). But on 1 certain dimension, naive O(n^2) can be O(n) by this well-known equation: sum[i..j] = sum[0..j] - sum[0..i]. And pls take care of several corner cases.

class Solution {
public:
/**
* @param matrix an integer matrix
* @return the coordinate of the left-up and right-down number
*/
vector<vector<int>> submatrixSum(vector<vector<int>>& m) { int h = m.size();
if(!h) return {};
int w = m[].size();
if(!w) return {}; // Get accumulate sum by columns
vector<vector<int>> cols(h, vector<int>(w, ));
for(int i = ; i < w; i ++)
{
unordered_map<int, int> rec; // sum-inx
for(int j = ; j < h; j ++)
{
if(m[j][i] == )
{
return {{i, j},{i, j}};
} cols[j][i] = (j ? cols[j - ][i] : ) + m[j][i];
if (!cols[j][i])
{
return {{, i}, {j, i}};
}
else if(rec.find(cols[j][i]) != rec.end())
{
return {{rec[cols[j][i]] + , i}, {j, i}};
}
rec[cols[j][i]] = j;
}
} // horizontal case
for(int i = ; i < h; i ++)
for(int j = i; j < h; j ++)
{
vector<int> hsum(w, );
for(int x = ; x < w; x ++)
{
int prev = ((i == ) ? : cols[i - ][x]);
hsum[x] = cols[j][x] - prev;
}
//
vector<int> asum(w, );
unordered_map<int, int> rec; // sum-inx
for(int x = ; x < w; x ++)
{
int nsum = (x ? asum[x - ] : ) + hsum[x];
if (!nsum)
{
return {{i + , }, {j, x}};
}
else if(rec.find(nsum) != rec.end())
{
return {{i, rec[nsum] + }, {j, x}};
}
rec[nsum] = x;
asum[x] = nsum;
}
} return {};
}
};

LintCode "Submatrix Sum"的更多相关文章

  1. [LintCode] Submatrix Sum 子矩阵之和

    Given an integer matrix, find a submatrix where the sum of numbers is zero. Your code should return ...

  2. array / matrix subarray/submatrix sum

    Maximal Subarray Sum : O(n) scan-and-update dynamic programming, https://en.wikipedia.org/wiki/Maxim ...

  3. lintcode 中等题:Submatrix sum is 0 和为零的子矩阵

    和为零的子矩阵 给定一个整数矩阵,请找出一个子矩阵,使得其数字之和等于0.输出答案时,请返回左上数字和右下数字的坐标. 样例 给定矩阵 [ [1 ,5 ,7], [3 ,7 ,-8], [4 ,-8 ...

  4. lintcode: k Sum 解题报告

    K SUM My Submissions http://www.lintcode.com/en/problem/k-sum/ 题目来自九章算法 13% Accepted Given n distinc ...

  5. LintCode "4 Sum"

    4 Pointer solution. Key: when moving pointers, we skip duplicated ones. Ref: https://github.com/xbz/ ...

  6. Submatrix Sum

    Given an integer matrix, find a submatrix where the sum of numbers is zero. Your code should return ...

  7. Lintcode: Subarray Sum 解题报告

    Subarray Sum 原题链接:http://lintcode.com/zh-cn/problem/subarray-sum/# Given an integer array, find a su ...

  8. LintCode Subarray Sum

    For this problem we need to learn a new trick that if your start sum up all elements in an array. Wh ...

  9. [LintCode] Two Sum 两数之和

    Given an array of integers, find two numbers such that they add up to a specific target number. The ...

随机推荐

  1. 'dict' object has no attribute 'a'

    a = {} #a.a = 'a' #AttributeError: 'dict' object has no attribute 'a' #a['a'] #KeyError: 'a' a['a'] ...

  2. Shell 条件表达式汇总

    条件表达式 文件表达式 if [ -f  file ]    如果文件存在if [ -d ...   ]    如果目录存在if [ -s file  ]    如果文件存在且非空 if [ -r f ...

  3. 319. Bulb Switche

    There are n bulbs that are initially off. You first turn on all the bulbs. Then, you turn off every ...

  4. Codeforces Round #365 (Div. 2) B 前缀和

    B. Mishka and trip time limit per test 1 second memory limit per test 256 megabytes input standard i ...

  5. Web上下文配置【MvcConfig】

    基于Servlet3.0规范和SpringMVC4注解式配置方式,实现零xml配置,弄了个小demo,供交流讨论. 项目说明如下: 1.db.sql是项目中用到的表,数据库使用的是oracle11g ...

  6. 卸载linux自带版本JDK

    1)卸载系统自带的jdk版本:    查看自带的jdk:    #rpm -qa|grep gcj    可能看到如下类似的信息:    libgcj-4.1.2-44.el5    java-1.4 ...

  7. 论文阅读之:PRIORITIZED EXPERIENCE REPLAY

    PRIORITIZED EXPERIENCE REPLAY ICLR 2016 经验回放使得 online reinforcement learning agent 能够记住并且回放过去的经验.在先前 ...

  8. 半透明背景(兼容IE)

    在CSS3中有rgba属性,可以很方便的实现背景透明,但对于IE家族来说却不是那么容易实现: FireFox.chrome.opera.safari 凡是对支持CSS3且支持W3标准的浏览器都可以现实 ...

  9. linux服务之snmp

    背景信息 http://datatracker.ietf.org/doc/rfc1213/ http://oid-info.com/get/1.3.6.1 http://oid-info.com/ge ...

  10. node-webkit 应用打包发布

    方便进行打包,使用了nodejs  ,gulp  nw-builder 备注  windows 操作系统部分版本需要包含  msvcr100.dll  建议制作安装程序的时候直接包含 为了进行视频以及 ...