算法题求解答：如何简化下面的代码？ js 代码

[DingDingDang123]

功能描述
当 const mergeTable=[2]时，
函数返回
{ rowSpan: 2 }
{ rowSpan: 0 }

当 const mergeTable=[2,3]时,函数返回
{ rowSpan: 2 }
{ rowSpan: 0 }
{ rowSpan: 3 }
{ rowSpan: 0 }
{ rowSpan: 0 }

当 const mergeTable=[2,3,3,2]函数返回
{ rowSpan: 2 }
{ rowSpan: 0 }
{ rowSpan: 3 }
{ rowSpan: 0 }
{ rowSpan: 0 }
{ rowSpan: 3 }
{ rowSpan: 0 }
{ rowSpan: 0 }
{ rowSpan: 2 }
{ rowSpan: 0 }

需要优化的函数
功能已经实现，代码冗余，请问如何优化


```
// const mergeTable=[2] // 测试 ok
const mergeTable = [2, 3]; // 测试 ok
// const mergeTable=[2,3,3,2]; // 测试 ok

// 如何简化这个函数
function renderRowSpan(index) {
for (let i in mergeTable) {
i=Number(i);
if (i === 0) {
if (index === 0) {
return { rowSpan: mergeTable[0] };
} else if (index < mergeTable[0]) {
return { rowSpan: 0 };
}
}

if (i === 1) {
if (index === mergeTable[0]) {
return { rowSpan: mergeTable[1] };
}
if (index < mergeTable[0] + mergeTable[1]) {
return { rowSpan: 0 };
}
}

if (i === 2) {
if (index === mergeTable[0] + mergeTable[1]) {
return { rowSpan: mergeTable[2] };
}
if (index < mergeTable[0] + mergeTable[1] + mergeTable[2]) {
return { rowSpan: 0 };
}
}


if (i === 3) {
if (index === mergeTable[0] + mergeTable[1]+mergeTable[2]) {
return { rowSpan: mergeTable[3] };
}
if (index < mergeTable[0] + mergeTable[1] + mergeTable[2]+ mergeTable[3]) {
return { rowSpan: 0 };
}
}
}
}

// 外部方法会渲染 index
for (let index = 0; index < 10; index++) {
console.log(renderRowSpan(index));
}

```

[MegrezZhu]

啥花里胡哨的

[binux]

function renderRowSpan(index) {
for (let i of mergeTable) {
if (index == 0) return { rowSpan: i };
if (index < 0) return { rowSpan: 0 };
index -= i;
}
if (index < 0) return { rowSpan: 0 };
}

[DingDingDang123]

@binux 大佬，太感谢了，测试可行，这就是我想要的效果

[DingDingDang123]

@MegrezZhu 感谢大佬，样式很漂亮，就是不能复制，不好测试。

[sweetcola]

const renderRowSpan = data => data.reduce((t, c) => ([...t, { rowSpan: c }, ...new Array(c - 1).fill({ rowSpan: 0 })]), []);

[Cbdy]

@DingDingDang123 升级一下 Monterey 就可以复制了

[des]

用 flat 更简洁易懂一点，顺便给你做了错误处理
const makeRowSpan = i => ({ rowSpan: i })
const makeRowBySpanNum = num => [makeRowSpan(num), ...new Array(--num).fill(makeRowSpan(0))]
const makeSpan = rows => rows?.filter(Number.isFinite).map(makeRowBySpanNum).flat() ?? []
makeSpan([2,3,3,null,2])