添加题72，322，92

Author: ligecarryme

基础算法篇/动态规划.md

@@ -367,5 +367,79 @@ function longestCommonSubsequence(text1: string, text2: string): number {
 };
 ```
 
+##### [72. 编辑距离](https://leetcode-cn.com/problems/edit-distance/)
 
+给你两个单词 `word1` 和 `word2`，请你计算出将 `word1` 转换成 `word2` 所使用的最少操作数 。
+
+```tsx
+function minDistance(word1: string, word2: string): number {
+    let l1: number = word1.length
+    let l2: number = word2.length
+    let dp: number[][] = new Array(l1 + 1).fill(0).map(() => new Array(l2 + 1).fill(0))
+    for (let i = 0; i <= l1; i++) {
+        dp[i][0] = i
+    }
+    for (let j = 0; j <= l2; j++) {
+        dp[0][j] = j
+    }
+    for (let i = 1; i <= l1; i++) {
+        for (let j = 1; j <= l2; j++) {
+            if (word1[i - 1] === word2[j - 1]) {
+                dp[i][j] = dp[i - 1][j - 1]
+            } else {
+                dp[i][j] = Math.min(dp[i - 1][j - 1], dp[i - 1][j], dp[i][j - 1]) + 1
+            }
+        }
+    }
+    return dp[l1][l2]
+};
+```
+
+## 零钱和背包（10%）
+
+##### [322. 零钱兑换](https://leetcode-cn.com/problems/coin-change/)
+
+给定不同面额的硬币 coins 和一个总金额 amount。编写一个函数来计算可以凑成总金额所需的最少的硬币个数。如果没有任何一种硬币组合能组成总金额，返回 -1。
+
+```tsx
+function coinChange(coins: number[], amount: number): number {
+    const dp: number[] = new Array(amount + 1).fill(amount + 1)
+    dp[0] = 0
+    for (let i = 1; i <= amount; i++) {
+        for (let j = 0; j < coins.length; j++) {
+            if (i - coins[j] >= 0) {
+                dp[i] = Math.min(dp[i], dp[i - coins[j]] + 1)
+            }
+        }
+    }
+    if (dp[amount] > amount) {
+        return -1
+    }
+    return dp[amount]
+};
+```
+
+##### [92.背包问题](https://www.lintcode.com/problem/backpack/description)
+
+在`n`个物品中挑选若干物品装入背包，最多能装多满？假设背包的大小为`m`，每个物品的大小为Ai
+
+```tsx
+function backpack(m: number, A: number[]): number {
+	const dp: boolean[][] = new Array(A.length+1).fill(false).map(()=>new Array(m+1).fill(false))
+	for(let i=1;i<=A.length;i++){
+        for(let j=0;j<=m;j++){
+            dp[i][j] = dp[i-1][j]
+            if(j-A[i-1] >= 0 && dp[i-1][j-A[i-1]]){
+                dp[i][j] = true
+            }
+        }
+    }
+    for(let i=m;i>=0;i--){
+        if(dp[A.length][i]){
+            return i
+        }
+    }
+    return 0
+}
+```