解法一:
#include
typedef struct {
int *elem;
int len
} SqList;
void MergeList(SqList *LA, SqList *LB, SqList *LC) {
int i, j, n;
for (i = LA->len - 1, j = LB->len - 1, n = i + j + 1; n >= 0;) {
LC->elem[n--] = (LA->elem[i] > LB->elem[j]) ? LA->elem[i--] : LB->elem[j--];
if (i == -1 || j == -1)
LC->elem[n--] = (i == -1) ? LB->elem[j--] : LA->elem[i--];
}
}
int main() {
int a[] = {-1, 1, 2, 5,25}, b[] = {-3, 0, 2, 10,30,35}, c[11];
SqList LA = {a, 5}, LB = {b, 6}, LC = {c, 11};
MergeList2(&LA, &LB, &LC);
for (int i = 0; i < 11; ++i) {
printf("%d\t", LC.elem[i]);
}
return 0;
}
解法二:
#include
typedef struct {
int *elem;
int len
} SqList;
void MergeList(SqList *LA, SqList *LB, SqList *LC) {
int as = 0, bs = 0, s = 0, i, j;
if (LA->len < LB->len) {
SqList *temp = LA;
LA = LB;
LB = temp;
}
for (i = 0; i < LA->len; i++) {
for (j = bs; j < LB->len; j++) {
if (LA->elem[i] < LB->elem[j]) {
LC->elem[s++] = LA->elem[i], as++;
break;
} else if (LA->elem[i] > LB->elem[j]) {
LC->elem[s++] = LB->elem[j], bs++;
} else {
LC->elem[s++] = LA->elem[i], LC->elem[s++] = LB->elem[j];
as++, bs++;
break;
}
}
}
if (bs == LB->len)
for (; as < LA->len; as++)
LC->elem[s++] = LA->elem[as];
}
int main() {
int a[] = {-1, 1, 2, 5,25}, b[] = {-3, 0, 2, 10,30,35}, c[11];
SqList LA = {a, 5}, LB = {b, 6}, LC = {c, 11};
MergeList2(&LA, &LB, &LC);
for (int i = 0; i < 11; ++i) {
printf("%d\t", LC.elem[i]);
}
return 0;
}
对解法一进行修改:
function mergeArray(arr1,arr2){
let arr3=[], i, j, n;
for (i = arr1.length - 1, j = arr2.length - 1, n = i + j + 1; n >= 0;) {
if (arr1[i] > arr2[j]) arr3[n--] = arr1[i--];
else arr3[n--] = arr2[j--];
if (i == -1 || j == -1) {
if (i == -1) arr3[n--] = arr2[j--];
else arr3[n--] = arr1[i--];
}
}
return arr3;
}
let arr1=[-1,1,2,6], arr2=[-3,0,2,10,12], arr3=mergeArray(arr1,arr2);
console.log(arr3);
对解法二进行修改:
function mergeArray(a, b) {
let c = [], as = 0, bs = 0, s = 0
if (a.length < b.length) [a, b] = [b, a] // 交换数组
for (let i = 0; i < a.length; i++) {
for (let j = bs; j < b.length; j++) {
if (a[i] < b[j]) {
c[s++] = a[i]
as++
break
} else if (a[i] > b[j]) {
c[s++] = b[j]
bs++
} else {
c[s++] = a[i]
c[s++] = b[j]
as++
bs++
break
}
}
}
if (bs == b.length)
for (; as < a.length; as++) c[s++] = a[as]
return c
}
let a = [-1, 1, 2, 5, 10, 12], b = [-3, 0, 2, 10, 12, 18, 20], c = mergeArray(a, b)
console.log(c)
在此感谢大佬提供的解法一代码~~