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93 lines
2.2 KiB
Markdown
Executable File
93 lines
2.2 KiB
Markdown
Executable File
# [64. Minimum Path Sum](https://leetcode.com/problems/minimum-path-sum/)
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## 题目
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Given a *m* x *n* grid filled with non-negative numbers, find a path from top left to bottom right which *minimizes* the sum of all numbers along its path.
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**Note**: You can only move either down or right at any point in time.
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**Example**:
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Input:
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[
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[1,3,1],
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[1,5,1],
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[4,2,1]
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]
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Output: 7
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Explanation: Because the path 1→3→1→1→1 minimizes the sum.
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## 题目大意
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给定一个包含非负整数的 m x n 网格，请找出一条从左上角到右下角的路径，使得路径上的数字总和为最小。说明：每次只能向下或者向右移动一步。
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## 解题思路
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- 在地图上求出从左上角到右下角的路径中，数字之和最小的一个，输出数字和。
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- 这一题最简单的想法就是用一个二维数组来 DP，当然这是最原始的做法。由于只能往下和往右走，只需要维护 2 列信息就可以了，从左边推到最右边即可得到最小的解。更近一步，可以直接在原来的数组中做原地 DP，空间复杂度为 0 。
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## 代码
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```go
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package leetcode
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// 解法一 原地 DP，无辅助空间
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func minPathSum(grid [][]int) int {
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m, n := len(grid), len(grid[0])
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for i := 1; i < m; i++ {
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grid[i][0] += grid[i-1][0]
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}
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for j := 1; j < n; j++ {
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grid[0][j] += grid[0][j-1]
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}
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for i := 1; i < m; i++ {
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for j := 1; j < n; j++ {
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grid[i][j] += min(grid[i-1][j], grid[i][j-1])
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}
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}
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return grid[m-1][n-1]
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}
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// 解法二 最原始的方法，辅助空间 O(n^2)
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func minPathSum1(grid [][]int) int {
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if len(grid) == 0 {
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return 0
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}
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m, n := len(grid), len(grid[0])
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if m == 0 || n == 0 {
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return 0
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}
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dp := make([][]int, m)
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for i := 0; i < m; i++ {
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dp[i] = make([]int, n)
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}
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// initFirstCol
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for i := 0; i < len(dp); i++ {
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if i == 0 {
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dp[i][0] = grid[i][0]
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} else {
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dp[i][0] = grid[i][0] + dp[i-1][0]
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}
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}
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// initFirstRow
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for i := 0; i < len(dp[0]); i++ {
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if i == 0 {
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dp[0][i] = grid[0][i]
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} else {
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dp[0][i] = grid[0][i] + dp[0][i-1]
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}
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}
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for i := 1; i < m; i++ {
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for j := 1; j < n; j++ {
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dp[i][j] = min(dp[i-1][j], dp[i][j-1]) + grid[i][j]
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}
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}
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return dp[m-1][n-1]
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}
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``` |