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算法(四)--二分法,动态规划

算法(四)--二分法,动态规划

作者: PurelightMe | 来源:发表于2021-12-18 14:48 被阅读0次

    二分法

    前提:

    • 有序
    • 上下界
    • 可以通过索引访问

    模板:

    left,right = 0,len(array)-1
    while left <= right:
        mid = (left+right)/2
        if array[mid] == target:
            break or return result
        elseif array[mid] < target:
        left = mid+1
      else:
        right = mid-1
    

    剑指 Offer II 072. 求平方根

    该题也可以用牛顿迭代法解题

    func mySqrt(x int) int {
        if x == 0 {
            return 0
        }
        left,right := 0,x
        for ;left <= right; {
          mid := (left+right)/2  //这里怕整形溢出的话,可以用: mid := left + (right-left)/2
            rs := mid*mid
            if rs == x {
                return mid
            }else if(rs < x){
                left = mid + 1
            }else{
                right = mid - 1
            }
        }
        return right
    }
    
    //牛顿迭代法
    

    367. 有效的完全平方数

    func isPerfectSquare(num int) bool {
        left,right := 0,num
        for ;left <= right; {
            mid := (left+right)/2
            if mid*mid == num {
                return true
            }else if mid*mid < num {
                left = mid + 1
            }else{
                right = mid - 1
            }
        }
        return false
    }
    

    33. 搜索旋转排序数组

    func search(nums []int, target int) int {
        left, right := 0, len(nums)-1
        for left <= right {
            mid := (right - left) / 2 + left
            if nums[mid] == target {
                return mid
            }
            if nums[mid] >= nums[left] {
                if nums[mid] > target && target >= nums[left] {
                    right = mid - 1
                } else {
                    left = mid + 1
                }
            } else {
                if nums[mid] < target && target <= nums[right] {
                    left = mid + 1
                } else {
                    right = mid - 1
                }
            }
        }
        return -1
    }
    

    74. 搜索二维矩阵

    func searchMatrix(matrix [][]int, target int) bool {
        m := len(matrix)
        if m == 0 {
            return false
        }
        n := len(matrix[0])
        for i := 0;i < m;i++ {
            if target > matrix[i][n-1] {
                continue
            }
            left,right := 0,n-1
            for ;left <= right; {
                mid := left + (right-left)/2
                if matrix[i][mid] == target {
                    return true
                }
                if matrix[i][mid] < target {
                    left = mid + 1
                }else{
                    right = mid - 1
                }
            }
        }
        return false
    }
    

    动态规划

    动态规划和递归,分治没有本质区别

    共性:找到重复子问题

    区别:是否有最优解

    62. 不同路径

    func uniquePaths(m int, n int) int {
        dp := make([][]int, m)
        for i := range dp {
            dp[i] = make([]int, n)
            dp[i][0] = 1
        }
        for j := 0; j < n; j++ {
            dp[0][j] = 1
        }
        for i := 1; i < m; i++ {
            for j := 1; j < n; j++ {
                dp[i][j] = dp[i-1][j] + dp[i][j-1]
            }
        }
        return dp[m-1][n-1]
    }
    //空间优化版本:
    func uniquePaths(m int, n int) int {
        cur := make([]int,n)
        for i :=0;i<n;i++{
            cur[i] = 1
        }
        for x := 1;x < m;x++ {
            for y := 1;y < n;y++ {
                cur[y] += cur[y-1]
            }
        }
        return cur[n-1]
    }
    //排列组合解法:
    func uniquePaths(m int, n int) int {
        return int(new(big.Int).Binomial(int64(m+n-2), int64(n-1)).Int64())
    }
    

    63. 不同路径 II

    func uniquePathsWithObstacles(obstacleGrid [][]int) int {
        n, m := len(obstacleGrid), len(obstacleGrid[0])
        f := make([]int, m)
        if obstacleGrid[0][0] == 0 {
            f[0] = 1
        }
        for i := 0; i < n; i++ {
            for j := 0; j < m; j++ {
                if obstacleGrid[i][j] == 1 {
                    f[j] = 0
                    continue
                }
                if j - 1 >= 0 && obstacleGrid[i][j-1] == 0 {
                    f[j] += f[j-1]
                }
            }
        }
        return f[len(f)-1]
    }
    

    1143. 最长公共子序列

    //DP方程
    if s1[-1] != s2[-1]
        LCS(s1,s2) = MAX(LCS(s1-1,s2),LCS(s1,s2-1))
    if s1[-1] == s2[-1]
        LCS[s1,s2] = LCS(s1-1,s2-1) + 1
    //代码:
    func longestCommonSubsequence(text1, text2 string) int {
        m, n := len(text1), len(text2)
        dp := make([][]int, m+1)
        for i := range dp {
            dp[i] = make([]int, n+1)
        }
        for i, c1 := range text1 {
            for j, c2 := range text2 {
                if c1 == c2 {
                    dp[i+1][j+1] = dp[i][j] + 1
                } else {
                    dp[i+1][j+1] = max(dp[i][j+1], dp[i+1][j])
                }
            }
        }
        return dp[m][n]
    }
    
    func max(a, b int) int {
        if a > b {
            return a
        }
        return b
    }
    

    120. 三角形最小路径和

    func minimumTotal(triangle [][]int) int {
        dp := triangle
        for i := len(triangle)-2;i >= 0;i-- {
            for j := len(triangle[i])-1;j >= 0;j-- {
                dp[i][j] += Min(dp[i+1][j],dp[i+1][j+1])
            }
        }
        return dp[0][0]
    }
    
    func Min(a,b int) int {
        if a > b {
            return b
        }
        return a
    }
    

    53. 最大子数组和

    //DP方程
    f(i) = Max(f(i-1),0) + a[i]
    
    func maxSubArray(nums []int) int {
        if len(nums) == 1 {
            return nums[0]
        }
        max := nums[0]
        for i := 1;i < len(nums);i++ {
            nums[i] = Max(nums[i],nums[i]+nums[i-1])
            max = Max(nums[i],max)
        }
        return max
    }
    
    func Max(a,b int) int {
        if a > b {
            return a
        }
        return b
    }
    

    152. 乘积最大子数组

    func maxProduct(nums []int) int {
        maxF, minF, ans := nums[0], nums[0], nums[0]
        for i := 1; i < len(nums); i++ {
            mx, mn := maxF, minF
            maxF = Max(mx * nums[i], Max(nums[i], mn * nums[i]))
            minF = Min(mn * nums[i], Min(nums[i], mx * nums[i]))
            ans = Max(maxF, ans)
        }
        return ans
    }
    
    func Max(a,b int) int {
        if a > b {
            return a
        }
        return b
    }
    
    func Min(a,b int) int {
        if a < b {
            return a
        }
        return b
    }
    

    322. 零钱兑换

    //DP
    func coinChange(coins []int, amount int) int {
        dp := make([]int, amount+1)
        dp[0] = 0
        for j := 1; j <= amount; j++ {
            dp[j] = math.MaxInt32
            for i := 0; i < len(coins); i++ {
                if j >= coins[i] && dp[j-coins[i]] != math.MaxInt32 {
                    dp[j] = min(dp[j], dp[j-coins[i]]+1)
                }
            }
        }
        if dp[amount] == math.MaxInt32 {
            return -1
        }
        return dp[amount]
    }
    
    func min(a, b int) int {
        if a < b {
            return a
        }
        return b
    }
    

    198. 打家劫舍

    func rob(nums []int) int {
        n := len(nums)
        if n == 0 {
            return 0
        }
        a := make([][]int,n)
        for i,_ := range a {
            a[i] = make([]int,2)
        }
        a[0][0] = 0
        a[0][1] = nums[0]
        for j := 1;j < n;j++ {
            a[j][0] = Max(a[j-1][0],a[j-1][1])
            a[j][1] = a[j-1][0] + nums[j]
        }
        return Max(a[n-1][0],a[n-1][1])
    }
    
    func Max(a,b int) int {
        if a > b {
            return a
        }
        return b
    }
    

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