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feat: add ts solution to lc problem: No.2137 #1724

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Sep 30, 2023
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Original file line number Diff line number Diff line change
Expand Up @@ -60,7 +60,7 @@

我们注意到,如果一个水量 $x$ 满足条件,那么所有小于 $x$ 的水量也满足条件。因此我们可以使用二分查找的方法找到最大的满足条件的水量。

我们定义二分查找的左边界 $l=0$,右边界 $r=\max(buckets)$。每次二分查找时,我们取 $l$ 和 $r$ 的中点 $m$,判断 $m$ 是否满足条件。如果满足条件,那么我们将 $l$ 更新为 $m$,否则我们将 $r$ 更新为 $m$。在二分查找结束后,最大的满足条件的水量即为 $l$。
我们定义二分查找的左边界 $l=0$,右边界 $r=\max(buckets)$。每次二分查找时,我们取 $l$ 和 $r$ 的中点 $mid$,判断 $mid$ 是否满足条件。如果满足条件,那么我们将 $l$ 更新为 $mid$,否则我们将 $r$ 更新为 $m$。在二分查找结束后,最大的满足条件的水量即为 $l$。

问题的关键转换为如果判断一个水量 $v$ 是否满足条件。我们可以遍历所有水桶,对于每个水桶,如果其水量大于 $v$,那么需要倒出 $x-v$ 的水量;如果其水量小于 $v$,那么需要向其中倒入 $(v-x)\times\frac{100}{100-\textit{loss}}$ 的水量。如果倒出的水量大于等于倒入的水量,那么说明 $v$ 满足条件。

Expand Down Expand Up @@ -195,7 +195,30 @@ func equalizeWater(buckets []int, loss int) float64 {
### **TypeScript**

```ts

function equalizeWater(buckets: number[], loss: number): number {
let l = 0;
let r = Math.max(...buckets);
const check = (v: number): boolean => {
let [a, b] = [0, 0];
for (const x of buckets) {
if (x >= v) {
a += x - v;
} else {
b += ((v - x) * 100) / (100 - loss);
}
}
return a >= b;
};
while (r - l > 1e-5) {
const mid = (l + r) / 2;
if (check(mid)) {
l = mid;
} else {
r = mid;
}
}
return l;
}
```

### **...**
Expand Down
Original file line number Diff line number Diff line change
Expand Up @@ -178,7 +178,30 @@ func equalizeWater(buckets []int, loss int) float64 {
### **TypeScript**

```ts

function equalizeWater(buckets: number[], loss: number): number {
let l = 0;
let r = Math.max(...buckets);
const check = (v: number): boolean => {
let [a, b] = [0, 0];
for (const x of buckets) {
if (x >= v) {
a += x - v;
} else {
b += ((v - x) * 100) / (100 - loss);
}
}
return a >= b;
};
while (r - l > 1e-5) {
const mid = (l + r) / 2;
if (check(mid)) {
l = mid;
} else {
r = mid;
}
}
return l;
}
```

### **...**
Expand Down
Original file line number Diff line number Diff line change
@@ -0,0 +1,24 @@
function equalizeWater(buckets: number[], loss: number): number {
let l = 0;
let r = Math.max(...buckets);
const check = (v: number): boolean => {
let [a, b] = [0, 0];
for (const x of buckets) {
if (x >= v) {
a += x - v;
} else {
b += ((v - x) * 100) / (100 - loss);
}
}
return a >= b;
};
while (r - l > 1e-5) {
const mid = (l + r) / 2;
if (check(mid)) {
l = mid;
} else {
r = mid;
}
}
return l;
}