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Commit ab46d51

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feat: update solutions to lc problem: No.0064
No.0064.Minimum Path Sum
1 parent 5524e1d commit ab46d51

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-13
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4 files changed

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‎solution/0000-0099/0064.Minimum Path Sum/README.md‎

Lines changed: 32 additions & 5 deletions
Original file line numberDiff line numberDiff line change
@@ -42,7 +42,9 @@
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<!-- 这里可写通用的实现逻辑 -->
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动态规划。假设 `dp[i][j]` 表示到达网格 `(i,j)` 的最小数字和,先初始化 dp 第一列和第一行的所有值,然后利用递推公式:`dp[i][j] = min(dp[i - 1][j], dp[i][j - 1]) + grid[i][j]` 求得 dp。
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**方法一:动态规划**
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假设 `dp[i][j]` 表示到达网格 `(i,j)` 的最小数字和,先初始化 dp 第一列和第一行的所有值,然后利用递推公式:`dp[i][j] = min(dp[i - 1][j], dp[i][j - 1]) + grid[i][j]` 求得 dp。
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最后返回 `dp[m - 1][n - 1]` 即可。
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@@ -107,15 +109,13 @@ function minPathSum(grid: number[][]): number {
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for (let j = 1; j < n; ++j) {
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dp[0][j] = dp[0][j - 1] + grid[0][j];
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}
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// dp
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for (let i = 1; i < m; ++i) {
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for (let j = 1; j < n; ++j) {
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let cur = grid[i][j];
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dp[i][j] = cur + Math.min(dp[i - 1][j], dp[i][j - 1]);
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dp[i][j] = Math.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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};
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```
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### **C++**
@@ -206,6 +206,33 @@ public class Solution {
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}
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```
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### **JavaScript**
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```js
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/**
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* @param {number[][]} grid
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* @return {number}
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*/
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var minPathSum = function (grid) {
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let m = grid.length,
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n = grid[0].length;
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let dp = Array.from({ length: m }, v => new Array(n).fill(0));
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dp[0][0] = grid[0][0];
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for (let i = 1; i < m; ++i) {
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dp[i][0] = dp[i - 1][0] + grid[i][0];
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}
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for (let j = 1; j < n; ++j) {
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dp[0][j] = dp[0][j - 1] + grid[0][j];
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}
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for (let i = 1; i < m; ++i) {
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for (let j = 1; j < n; ++j) {
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dp[i][j] = Math.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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```
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### **...**
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```

‎solution/0000-0099/0064.Minimum Path Sum/README_EN.md‎

Lines changed: 29 additions & 4 deletions
Original file line numberDiff line numberDiff line change
@@ -95,15 +95,13 @@ function minPathSum(grid: number[][]): number {
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for (let j = 1; j < n; ++j) {
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dp[0][j] = dp[0][j - 1] + grid[0][j];
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}
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// dp
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for (let i = 1; i < m; ++i) {
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for (let j = 1; j < n; ++j) {
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let cur = grid[i][j];
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dp[i][j] = cur + Math.min(dp[i - 1][j], dp[i][j - 1]);
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dp[i][j] = Math.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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};
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```
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### **C++**
@@ -194,6 +192,33 @@ public class Solution {
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}
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```
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### **JavaScript**
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```js
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/**
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* @param {number[][]} grid
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* @return {number}
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*/
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var minPathSum = function (grid) {
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let m = grid.length,
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n = grid[0].length;
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let dp = Array.from({ length: m }, v => new Array(n).fill(0));
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dp[0][0] = grid[0][0];
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for (let i = 1; i < m; ++i) {
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dp[i][0] = dp[i - 1][0] + grid[i][0];
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}
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for (let j = 1; j < n; ++j) {
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dp[0][j] = dp[0][j - 1] + grid[0][j];
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}
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for (let i = 1; i < m; ++i) {
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for (let j = 1; j < n; ++j) {
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dp[i][j] = Math.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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```
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### **...**
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```
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Original file line numberDiff line numberDiff line change
@@ -0,0 +1,22 @@
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/**
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* @param {number[][]} grid
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* @return {number}
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*/
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var minPathSum = function (grid) {
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let m = grid.length,
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n = grid[0].length;
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let dp = Array.from({ length: m }, v => new Array(n).fill(0));
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dp[0][0] = grid[0][0];
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for (let i = 1; i < m; ++i) {
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dp[i][0] = dp[i - 1][0] + grid[i][0];
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}
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for (let j = 1; j < n; ++j) {
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dp[0][j] = dp[0][j - 1] + grid[0][j];
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}
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for (let i = 1; i < m; ++i) {
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for (let j = 1; j < n; ++j) {
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dp[i][j] = Math.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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};

‎solution/0000-0099/0064.Minimum Path Sum/Solution.ts‎

Lines changed: 2 additions & 4 deletions
Original file line numberDiff line numberDiff line change
@@ -9,12 +9,10 @@ function minPathSum(grid: number[][]): number {
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for (let j = 1; j < n; ++j) {
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dp[0][j] = dp[0][j - 1] + grid[0][j];
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}
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// dp
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for (let i = 1; i < m; ++i) {
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for (let j = 1; j < n; ++j) {
15-
let cur = grid[i][j];
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dp[i][j] = cur + Math.min(dp[i - 1][j], dp[i][j - 1]);
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dp[i][j] = Math.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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};

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