本问题对应的 leetcode 原文链接：剑指 Offer 47. 礼物的最大价值
问题描述
在一个 m*n 的棋盘的每一格都放有一个礼物，每个礼物都有一定的价值（价值大于 0）。你可以从棋盘的左上角开始拿格子里的礼物，并每次向右或者向下移动一格、直到到达棋盘的右下角。给定一个棋盘及其上面的礼物的价值，请计算你最多能拿到多少价值的礼物？
示例 1:
输入:
[
[1,3,1],
[1,5,1],
[4,2,1]
]
输出: 12
解释: 路径 1→3→5→2→1 可以拿到最多价值的礼物
提示:
0 < grid.length <= 2000 < grid[0].length <= 200
解题思路
视频讲解直达： 本题视频讲解
代码实现
class Solution {
public int maxValue(int[][] grid) {
//优化版本
int n = grid.length;
int m = grid[0].length;
int[]dp = new int[m];
dp[0] = grid[0][0];
for(int j = 1; j < m; j++){
dp[j] = dp[j-1] + grid[0][j];
}
for(int i = 1; i < n; i++){
//
dp[0] = dp[0] + grid[i][0];
for(int j = 1; j < m; j++){
dp[j] = Math.max(dp[j], dp[j-1]) + grid[i][j];
}
}
return dp[m-1];
}
// 二维版本
public int maxValue(int[][] grid) {
int n = grid.length;
int m = grid[0].length;
int[][] dp = new int[n][m];
dp[0][0] = grid[0][0];
for(int i = 1; i < n; i++){
dp[i][0] = dp[i-1][0] + grid[i][0];
}
for(int j = 1; j < m; j++){
dp[0][j] = dp[0][j -1] + grid[0][j];
}
for(int i = 1; i < n; i++){
for(int j = 1; j < m; j++){
dp[i][j] = Math.max(dp[i-1][j], dp[i][j-1]) + grid[i][j];
}
}
return dp[n-1][m-1];
}
}
时间复杂度：O(n*m)
空间复杂度：O(n)
Python
class Solution(object):
def maxValue(self, grid):
"""
:type grid: List[List[int]]
:rtype: int
"""
n = len(grid)
m = len(grid[0])
dp = [0] * m
dp[0] = grid[0][0]
for j in range(1, m):
dp[j] = dp[j-1] + grid[0][j]
for i in range(1, n):
dp[0] = dp[0] + grid[i][0]
for j in range(1, m):
dp[j] = max(dp[j], dp[j-1]) + grid[i][j]
return dp[m-1]
C++
class Solution {
public:
int maxValue(vector<vector<int>>& grid) {
int n = grid.size();
int m = grid[0].size();
vector<int> dp(m, 0);
dp[0] = grid[0][0];
for (int j = 1; j < m; j++) {
dp[j] = dp[j-1] + grid[0][j];
}
for (int i = 1; i < n; i++) {
dp[0] += grid[i][0];
for (int j = 1; j < m; j++) {
dp[j] = max(dp[j-1], dp[j]) + grid[i][j];
}
}
return dp[m-1];
}
};
Go
func maxValue(grid [][]int) int {
n, m := len(grid), len(grid[0])
dp := make([]int, m)
dp[0] = grid[0][0]
for j := 1; j < m; j++ {
dp[j] = dp[j-1] + grid[0][j]
}
for i := 1; i < n; i++ {
dp[0] += grid[i][0]
for j := 1; j < m; j++ {
dp[j] = max(dp[j-1], dp[j]) + grid[i][j]
}
}
return dp[m-1]
}
func maxValue2(grid [][]int) int {
n, m := len(grid), len(grid[0])
dp := make([][]int, n)
for i := range dp {
dp[i] = make([]int, m)
}
dp[0][0] = grid[0][0]
for i := 1; i < n; i++ {
dp[i][0] = dp[i-1][0] + grid[i][0]
}
for j := 1; j < m; j++ {
dp[0][j] = dp[0][j-1] + grid[0][j]
}
for i := 1; i < n; i++ {
for j := 1; j < m; j++ {
dp[i][j] = max(dp[i-1][j], dp[i][j-1]) + grid[i][j]
}
}
return dp[n-1][m-1]
}
func max(a, b int) int {
if a > b {
return a
}
return b
}
JS
/**
* @param {number[][]} grid
* @return {number}
*/
var maxValue = function(grid) {
// 优化版本
let n = grid.length;
let m = grid[0].length;
let dp = new Array(m).fill(0);
dp[0] = grid[0][0];
for (let j = 1; j < m; j++) {
dp[j] = dp[j - 1] + grid[0][j];
}
for (let i = 1; i < n; i++) {
dp[0] = dp[0] + grid[i][0];
for (let j = 1; j < m; j++) {
dp[j] = Math.max(dp[j], dp[j - 1]) + grid[i][j];
}
}
return dp[m - 1];
// 二维版本
/*
let n = grid.length;
let m = grid[0].length;
let dp = new Array(n);
for (let i = 0; i < n; i++) {
dp[i] = new Array(m).fill(0);
}
dp[0][0] = grid[0][0];
for (let i = 1; i < n; i++) {
dp[i][0] = dp[i - 1][0] + grid[i][0];
}
for (let j = 1; j < m; j++) {
dp[0][j] = dp[0][j - 1] + grid[0][j];
}
for (let i = 1; i < n; i++) {
for (let j = 1; j < m; j++) {
dp[i][j] = Math.max(dp[i - 1][j], dp[i][j - 1]) + grid[i][j];
}
}
return dp[n - 1][m - 1];
*/
};
shuaidi