2021-12-28：给定一个二维数组matrix，matrix[i][j] = k代表: 从(i,j)位置可以随意往右跳＜=k步，或者从(i,j)位置可以随意往下跳＜=k步， 如果matrix[i]

Posted 2022-02-03 福大大架构师每日一题

2021-12-28：给定一个二维数组matrix，matrix[i][j] = k代表:
从(i,j)位置可以随意往右跳<=k步，或者从(i,j)位置可以随意往下跳<=k步，
如果matrix[i][j] = 0，代表来到(i,j)位置必须停止，
返回从matrix左上角到右下角，至少要跳几次，
已知matrix中行数n <= 5000, 列数m <= 5000，
matrix中的值，<= 5000。
来自京东。

答案2021-12-28：

方法一：自然智慧。递归。复杂度过不了。
方法二：动态规划+线段树。

代码用golang编写。代码如下：

package main

import (
    "fmt"
    "math"
)

func main() 
    ret := jump2([][]int1, 2, 3, 4, 5, 6, 7, 8, 9, 10)
    fmt.Println(ret)


func getMin(a, b int) int 
    if a < b 
        return a
     else 
        return b
    


// 优化方法, 利用线段树做枚举优化
// 因为线段树，下标从1开始
// 所以，该方法中所有的下标，请都从1开始，防止乱！
func jump2(arr [][]int) int 
    n := len(arr)
    m := len(arr[0])
    map0 := make([][]int, n+1)
    for i := 0; i < n+1; i++ 
        map0[i] = make([]int, m+1)
    
    for a, b := 0, 1; a < n; a, b = a+1, b+1 
        for c, d := 0, 1; c < m; c, d = c+1, d+1 
            map0[b][d] = arr[a][c]
        
    
    rowTrees := make([]*SegmentTree, n+1)
    for i := 1; i <= n; i++ 
        rowTrees[i] = NewSegmentTree(m)
    
    colTrees := make([]*SegmentTree, m+1)
    for i := 1; i <= m; i++ 
        colTrees[i] = NewSegmentTree(n)
    
    rowTrees[n].update0(m, m, 0, 1, m, 1)
    colTrees[m].update0(n, n, 0, 1, n, 1)
    for col := m - 1; col >= 1; col-- 
        if map0[n][col] != 0 
            left := col + 1
            right := getMin(col+map0[n][col], m)
            next := rowTrees[n].query(left, right, 1, m, 1)
            if next != math.MaxInt64 
                rowTrees[n].update0(col, col, next+1, 1, m, 1)
                colTrees[col].update0(n, n, next+1, 1, n, 1)
            
        
    
    for row := n - 1; row >= 1; row-- 
        if map0[row][m] != 0 
            up := row + 1
            down := getMin(row+map0[row][m], n)
            next := colTrees[m].query(up, down, 1, n, 1)
            if next != math.MaxInt64 
                rowTrees[row].update0(m, m, next+1, 1, m, 1)
                colTrees[m].update0(row, row, next+1, 1, n, 1)
            
        
    
    for row := n - 1; row >= 1; row-- 
        for col := m - 1; col >= 1; col-- 
            if map0[row][col] != 0 
                // (row,col) 往右是什么范围呢？[left,right]
                left := col + 1
                right := getMin(col+map0[row][col], m)
                next1 := rowTrees[row].query(left, right, 1, m, 1)
                // (row,col) 往下是什么范围呢？[up,down]
                up := row + 1
                down := getMin(row+map0[row][col], n)
                next2 := colTrees[col].query(up, down, 1, n, 1)
                next := getMin(next1, next2)
                if next != math.MaxInt64 
                    rowTrees[row].update0(col, col, next+1, 1, m, 1)
                    colTrees[col].update0(row, row, next+1, 1, n, 1)
                
            
        
    
    return rowTrees[1].query(1, 1, 1, m, 1)


// 区间查询最小值的线段树
// 注意下标从1开始，不从0开始
// 比如你传入size = 8
// 则位置对应为1~8，而不是0~7
type SegmentTree struct 
    min    []int
    change []int
    update []bool


func NewSegmentTree(size int) *SegmentTree 
    ret := &SegmentTree
    N := size + 1
    ret.min = make([]int, N<<2)
    ret.change = make([]int, N<<2)
    ret.update = make([]bool, N<<2)
    ret.update0(1, size, math.MaxInt64, 1, size, 1)
    return ret


func (this *SegmentTree) pushUp(rt int) 
    this.min[rt] = getMin(this.min[rt<<1], this.min[rt<<1|1])


func (this *SegmentTree) pushDown(rt, ln, rn int) 
    if this.update[rt] 
        this.update[rt<<1] = true
        this.update[rt<<1|1] = true
        this.change[rt<<1] = this.change[rt]
        this.change[rt<<1|1] = this.change[rt]
        this.min[rt<<1] = this.change[rt]
        this.min[rt<<1|1] = this.change[rt]
        this.update[rt] = false
    


// 最后三个参数是固定的, 每次传入相同的值即可:
// l = 1(固定)
// r = size(你设置的线段树大小)
// rt = 1(固定)
func (this *SegmentTree) update0(L, R, C, l, r, rt int) 
    if L <= l && r <= R 
        this.update[rt] = true
        this.change[rt] = C
        this.min[rt] = C
        return
    
    mid := (l + r) >> 1
    this.pushDown(rt, mid-l+1, r-mid)
    if L <= mid 
        this.update0(L, R, C, l, mid, rt<<1)
    
    if R > mid 
        this.update0(L, R, C, mid+1, r, rt<<1|1)
    
    this.pushUp(rt)


// 最后三个参数是固定的, 每次传入相同的值即可:
// l = 1(固定)
// r = size(你设置的线段树大小)
// rt = 1(固定)
func (this *SegmentTree) query(L, R, l, r, rt int) int 
    if L <= l && r <= R 
        return this.min[rt]
    
    mid := (l + r) >> 1
    this.pushDown(rt, mid-l+1, r-mid)
    left := math.MaxInt64
    right := math.MaxInt64
    if L <= mid 
        left = this.query(L, R, l, mid, rt<<1)
    
    if R > mid 
        right = this.query(L, R, mid+1, r, rt<<1|1)
    
    return getMin(left, right)

执行结果如下：

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左神java代码