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([][]int{{1, 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：

***

[左神java代码](