标签：线性 rgs res code stat bsp 经典 合并 int
一.总述
分治算法其实就是将一个大问题分解为若干个类型相同但是规模较小的子问题，使用递归的方式一直分解下去，然后将子问题的解合并得到原问题的解的策略。
二.经典的分治算法列举
二分搜索、大整数乘法、strassen矩阵乘法、棋盘覆盖、合并排序、快速排序、线性时间选择、最接近点对问题、循环赛日程表、汉诺塔等
三.分治算法举例
最大子数列问题：下面是股票的价格变动，求在哪一天买入，哪一天卖出获得的收益最高
1.双重循环实现（不是分治算法）
1）C#实现及结果
int[] prices = { 100, 113, 110, 85, 105, 102, 86, 63, 81, 101, 94, 106, 101, 79, 94, 90, 97 }; int profit = prices[1] - prices[0]; int tempProfit = 0; int[] result = { 0, 1 }; for(int i = 0;i < prices.Length;i++) { for(int j = i;j < prices.Length; j++) { tempProfit = prices[j] - prices[i]; if(profit < tempProfit) { profit = tempProfit; result[0] = i; result[1] = j; } } } Console.WriteLine("在第" + result[0] + "天买入，第" + result[1] + "天卖出，所获得利润最多，每股利润是" + profit + "元");
2）Lua实现代码及结果
prices = { 100, 113, 110, 85, 105, 102, 86, 63, 81, 101, 94, 106, 101, 79, 94, 90, 97 }
profit = prices[2] - prices[1]
tempProfit = 0
result = { 0, 1 }
for i=1,#prices,1 do
for j=i,#prices,1 do
tempProfit = prices[j] - prices[i];
if (profit < tempProfit) then
profit = tempProfit
result[0] = i-1
result[1] = j-1
end
end
end
print("在第"..result[0].."天买入，第"..result[1].."天卖出，所获得利润最多，每股利润是"..profit.."元")
2.分治算法
1）c#实现
static void Main(string[] args) { int[] prices = { 100, 113, 110, 85, 105, 102, 86, 63, 81, 101, 94, 106, 101, 79, 94, 90, 97 }; int start = 0; int end = prices.Length - 1; int profit = GetMaxProfit(prices,ref start,ref end); Console.WriteLine("在第" + start + "天买入，第" + end + "天卖出，所获得利润最多，每股利润是" + profit + "元"); } static int MaxNumInArray(int[] nums,int start,int end,out int maxNumIndex) { int result = nums[start]; maxNumIndex = start; for(int i = start +1;i <= end; i++) { if (nums[i] > result) { result = nums[i]; maxNumIndex = i; } } return result; } static int MinNumInArray(int[] nums,int start,int end,out int minNumIndex) { int result = nums[start]; minNumIndex = start; for (int i = start + 1; i <= end; i++) { if (nums[i] < result) { result = nums[i]; minNumIndex = i; } } return result; } static int GetMaxProfit(int[] nums,ref int start,ref int end) { if (start == end) return 0; int mid = (start + end) / 2; int crossStart; int crossEnd; int crossProfit = MaxNumInArray(nums,mid + 1,end,out crossEnd) - MinNumInArray(nums,start,mid,out crossStart); int leftStart = start; int leftEnd = mid; int leftProfit = GetMaxProfit(nums,ref leftStart,ref leftEnd); int rightStart = mid + 1; int rightEnd = end; int rightProfit = GetMaxProfit(nums,ref rightStart,ref rightEnd); int result; if(crossProfit > leftProfit && crossProfit > rightProfit) { result = crossProfit; start = crossStart; end = crossEnd; } else if(leftProfit > crossProfit && leftProfit > rightProfit) { result = leftProfit; start = leftStart; end = leftEnd; } else { result = rightProfit; start = rightStart; end = rightEnd; } return result; }
2）.Lua实现
prices = { 100, 113, 110, 85, 105, 102, 86, 63, 81, 101, 94, 106, 101, 79, 94, 90, 97 }
function MaxNumInArray(nums,smax,emax)
resultmax = nums[smax]
maxNumIndex = smax
for imax = smax,emax,1 do
if (nums[imax] > resultmax) then
resultmax = nums[imax]
maxNumIndex = imax
end
end
return resultmax,maxNumIndex;
end
function MinNumInArray(nums,smin,emin)
resultmin = nums[smin]
minNumIndex = smin
for i = smin +1,emin,1 do
if (nums[i] < resultmin) then
resultmin = nums[i]
minNumIndex = i
end
end
return resultmin,minNumIndex
end
function GetMaxProfit(nums,s,e)
if s==e then
return 1,s,e
end
local mid = 0;
if (s + e) % 2 == 0 then
mid = (s + e) / 2
else
mid = (s + e - 1) / 2
end
local rightMax,crossEnd = MaxNumInArray(nums,mid + 1,e)
local leftMin,crossStart = MinNumInArray(nums,s,mid)
local crossProfit = rightMax - leftMin
local leftProfit,leftStart,leftEnd = GetMaxProfit(nums,s,mid)
local rightProfit,rightStart,rightEnd = GetMaxProfit(nums,mid+1,e)
if crossProfit > leftProfit and crossProfit > rightProfit then
result = crossProfit
s = crossStart
e = crossEnd
elseif (leftProfit > crossProfit and leftProfit > rightProfit) then
result = leftProfit
s = leftStart
e = leftEnd
else
result = rightProfit;
s = rightStart;
e = rightEnd;
end
return result,s,e
end
profit,s,e = GetMaxProfit(prices,1,#prices)
print("在第"..(s-1).."天买入，第"..(e-1).."天卖出，所获得利润最多，每股利润是"..profit.."元")
标签：线性 rgs res code stat bsp 经典 合并 int
原文地址：https://www.cnblogs.com/movin2333/p/14402951.html
