Merge pull request #1283 from xiaofei-2020/dp17

添加（0474.一和零.md）：增加typescript版本

problems/0474.一和零.md

@@ -323,6 +323,129 @@ const findMaxForm = (strs, m, n) => {
 };
 ```
 
+TypeScript：
+
+> 滚动数组，二维数组法
+
+```typescript
+type BinaryInfo = { numOfZero: number, numOfOne: number };
+function findMaxForm(strs: string[], m: number, n: number): number {
+    const goodsNum: number = strs.length;
+    const dp: number[][] = new Array(m + 1).fill(0)
+        .map(_ => new Array(n + 1).fill(0));
+    for (let i = 0; i < goodsNum; i++) {
+        const { numOfZero, numOfOne } = countBinary(strs[i]);
+        for (let j = m; j >= numOfZero; j--) {
+            for (let k = n; k >= numOfOne; k--) {
+                dp[j][k] = Math.max(dp[j][k], dp[j - numOfZero][k - numOfOne] + 1);
+            }
+        }
+    }
+    return dp[m][n];
+};
+function countBinary(str: string): BinaryInfo {
+    let numOfZero: number = 0,
+        numOfOne: number = 0;
+    for (let s of str) {
+        if (s === '0') {
+            numOfZero++;
+        } else {
+            numOfOne++;
+        }
+    }
+    return { numOfZero, numOfOne };
+}
+```
+
+> 传统背包，三维数组法
+
+```typescript
+type BinaryInfo = { numOfZero: number, numOfOne: number };
+function findMaxForm(strs: string[], m: number, n: number): number {
+    /**
+        dp[i][j][k]: 前i个物品中, 背包的0容量为j, 1容量为k, 最多能放的物品数量
+     */
+    const goodsNum: number = strs.length;
+    const dp: number[][][] = new Array(goodsNum).fill(0)
+        .map(_ => new Array(m + 1)
+            .fill(0)
+            .map(_ => new Array(n + 1).fill(0))
+        );
+    const { numOfZero, numOfOne } = countBinary(strs[0]);
+    for (let i = numOfZero; i <= m; i++) {
+        for (let j = numOfOne; j <= n; j++) {
+            dp[0][i][j] = 1;
+        }
+    }
+    for (let i = 1; i < goodsNum; i++) {
+        const { numOfZero, numOfOne } = countBinary(strs[i]);
+        for (let j = 0; j <= m; j++) {
+            for (let k = 0; k <= n; k++) {
+                if (j < numOfZero || k < numOfOne) {
+                    dp[i][j][k] = dp[i - 1][j][k];
+                } else {
+                    dp[i][j][k] = Math.max(dp[i - 1][j][k], dp[i - 1][j - numOfZero][k - numOfOne] + 1);
+                }
+            }
+        }
+    }
+    return dp[dp.length - 1][m][n];
+};
+function countBinary(str: string): BinaryInfo {
+    let numOfZero: number = 0,
+        numOfOne: number = 0;
+    for (let s of str) {
+        if (s === '0') {
+            numOfZero++;
+        } else {
+            numOfOne++;
+        }
+    }
+    return { numOfZero, numOfOne };
+}
+```
+
+> 回溯法（会超时）
+
+```typescript
+function findMaxForm(strs: string[], m: number, n: number): number {
+    /**
+        思路：暴力枚举strs的所有子集，记录符合条件子集的最大长度
+     */
+    let resMax: number = 0;
+    backTrack(strs, m, n, 0, []);
+    return resMax;
+    function backTrack(
+        strs: string[], m: number, n: number,
+        startIndex: number, route: string[]
+    ): void {
+        if (startIndex === strs.length) return;
+        for (let i = startIndex, length = strs.length; i < length; i++) {
+            route.push(strs[i]);
+            if (isValidSubSet(route, m, n)) {
+                resMax = Math.max(resMax, route.length);
+                backTrack(strs, m, n, i + 1, route);
+            }
+            route.pop();
+        }
+    }
+};
+function isValidSubSet(strs: string[], m: number, n: number): boolean {
+    let zeroNum: number = 0,
+        oneNum: number = 0;
+    strs.forEach(str => {
+        for (let s of str) {
+            if (s === '0') {
+                zeroNum++;
+            } else {
+                oneNum++;
+            }
+        }
+    });
+    return zeroNum <= m && oneNum <= n;
+}
+```
+
 
 
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